What is the formula to convert °F to °C?  a) °C = °F + 273  b) °C = 5/9 (°F - 32)  c) °C = 9/5 (°F)+32  d) None of the above

What is the formula to convert °F to
°C?
 a) °C = °F + 273
 b) °C = 5/9 (°F - 32)
 c) °C = 9/5 (°F)+32
 d) None of the above
by

1 Answer

Answer :

°C = 5/9 (°F - 32)
by

Related questions

What is the formula to convert °C to °F?  a) °F = °C + 273  b) °F = 5/9 (°C - 32)  c) °F = 9/5 (°C)+32  d) None of the above

Description : What is the formula to convert °C to °F?  a) °F = °C + 273  b) °F = 5/9 (°C - 32)  c) °F = 9/5 (°C)+32  d) None of the above

Last Answer : momentum

A System has a temperature of 250°F. Convert this Value to °R?  a. 740°R  b.730°R  c. 720°R  d. 710°R formula: °R= °F + 460

Description : A System has a temperature of 250°F. Convert this Value to °R?  a. 740°R  b.730°R  c. 720°R  d. 710°R formula: °R= °F + 460

Last Answer : 710°R

212 °F = _____ °C  a. 200  b. 150  c. 100  d. None of the above

Description : 212 °F = _____ °C  a. 200  b. 150  c. 100  d. None of the above

Last Answer : 100

What is the SI unit of specific heat capacity?  A. J/kg  B. J/kg °F ∙  C. J/kg °C ∙  D. J/°C

Description : What is the SI unit of specific heat capacity?  A. J/kg  B. J/kg °F ∙  C. J/kg °C ∙  D. J/°C

Last Answer : J/kg °C

Convert the change of temperature from 20˚C to 30˚C to Kelvin scale.  a. 10 K  b. 293 K  c. 303 K  d. 273 K

Description : Convert the change of temperature from 20˚C to 30˚C to Kelvin scale.  a. 10 K  b. 293 K  c. 303 K  d. 273 K

Last Answer : 10 K

Utilizing the answer to the previous problem, estimate the overall or average increase in temperature ( ΔT) of the concrete roof from the energy absorbed from the sun during a12hour day. Assume that all of the radiation absorbed goes into heating the roof. The specific heat of concrete is about 900 J/kg, and the density is about 2,300 kg/m3 .  a. 7.9 °C  b. 8.9°C  c. 9.9°C  d. 10.9°C formula: ΔQ = m c ΔT

Description : Utilizing the answer to the previous problem, estimate the overall or average increase in temperature ( ΔT) of the concrete roof from the energy absorbed from the sun during a12hour day. Assume that all of the radiation absorbed goes into ... °C  b. 8.9°C  c. 9.9°C  d. 10.9°C formula: ΔQ = m c ΔT

Last Answer : 7.9 °C

212 °F = _____R  a. 567  b. 672  c. 700  d. None of the above

Description : 212 °F = _____R  a. 567  b. 672  c. 700  d. None of the above

Last Answer : 672

876 °R = _____ °F  a. 335  b. 416  c. 400  d. None of the above

Description : 876 °R = _____ °F  a. 335  b. 416  c. 400  d. None of the above

Last Answer : 416

746 °R = ______ °F  a. 254  b. 345  c. 286  d. None of the above

Description : 746 °R = ______ °F  a. 254  b. 345  c. 286  d. None of the above

Last Answer : 286

The condition of perfect vacuum, i.e., absolute zero pressure can be attained at  (a) a temperature of – 273.16°C  (b) a temperature of 0°C  (c) a temperature of 273 °K  (d) a negative pressure and 0°C temperature  (e) can’t be attained.

Description : The condition of perfect vacuum, i.e., absolute zero pressure can be attained at  (a) a temperature of – 273.16°C  (b) a temperature of 0°C  (c) a temperature of 273 °K  (d) a negative pressure and 0°C temperature  (e) can’t be attained.

Last Answer : Answer : a

Absolute zero pressure will occur  (a) at sea level  (b) at the center of the earth  (c) when molecular momentum of the system becomes zero  (d) under vacuum conditions  (e) at a temperature of – 273 °K

Description : Absolute zero pressure will occur  (a) at sea level  (b) at the center of the earth  (c) when molecular momentum of the system becomes zero  (d) under vacuum conditions  (e) at a temperature of – 273 °K

Last Answer : Answer : c

No liquid can exist as liquid at  (a) – 273 °K  (b) vacuum  (c) zero pressure  (d) centre of earth  (e) in space.

Description : No liquid can exist as liquid at  (a) – 273 °K  (b) vacuum  (c) zero pressure  (d) centre of earth  (e) in space.

Last Answer : Answer : c

A perfect gas has a value of R= 319.2 J/ kf.K and k= 1.26. If 120 kJ are added to 2.27 kf\g of this gas at constant pressure when the initial temp is 32.2°C? Find T2.  a. 339.4 K  b. 449.4 K  c. 559.4K  d. 669.4K formula: cp = kR/ k-1 Q= mcp(T2-T1)

Description : A perfect gas has a value of R= 319.2 J/ kf.K and k= 1.26. If 120 kJ are added to 2.27 kf\g of this gas at constant pressure when the initial temp is 32.2°C? Find T2.  a. 339.4 K  b. 449.4 K  c. 559.4K  d. 669.4K formula: cp = kR/ k-1 Q= mcp(T2-T1)

Last Answer : 339.4 K

Twenty grams of oxygen gas are compressed at a constant temperature of 30 ˚C to 5%of their original volume. What work is done on the system.  A.824 cal  B.924 cal  C.944 cal  D.1124 cal Formula: W = -mRTln (V2/V1) Where R = (1.98 cal/gmole·K) (32 g/gmole)

Description : Twenty grams of oxygen gas are compressed at a constant temperature of 30 ˚C to 5%of their original volume. What work is done on the system.  A.824 cal  B.924 cal  C.944 cal  D.1124 cal Formula: W = -mRTln (V2/V1) Where R = (1.98 cal/gmole·K) (32 g/gmole)

Last Answer : 1124 cal

212 °C = _____ K  a. 485  b. 435  c. 498  d. None of the above

Description : 212 °C = _____ K  a. 485  b. 435  c. 498  d. None of the above

Last Answer : 485

710°R= ______ °C  a. 214  b. 121  c. 213  d. None of the above

Description : 710°R= ______ °C  a. 214  b. 121  c. 213  d. None of the above

Last Answer : 121

What is the specific heat capacity of water in J/kg °C? ∙  A. 4581  B. 4185  C. 4518  D. 4815

Description : What is the specific heat capacity of water in J/kg °C? ∙  A. 4581  B. 4185  C. 4518  D. 4815

Last Answer : 4185

The amount of heat required to raise the temperature of 1kg of water through 1 °C is called ______.  A. Calorie  B. Joule  C. BTU  D. Kilocalorie

Description : The amount of heat required to raise the temperature of 1kg of water through 1 °C is called ______.  A. Calorie  B. Joule  C. BTU  D. Kilocalorie

Last Answer : Kilocalorie

The unit’of universal gas constant is  (a) watts/°K  (b) dynes/°C  (c) ergscm/°K  (d)erg/°K  (e) none of the above.

Description : The unit’of universal gas constant is  (a) watts/°K  (b) dynes/°C  (c) ergscm/°K  (d)erg/°K  (e) none of the above.

Last Answer : Answer : d

Find the change in internal energy of 5 lb. of oxygen gas when the temperature changes from 100 ˚F to 120 ˚F. CV = 0.157 BTU/lbm-˚R  A.14.7 BTU  B.15.7 BTU  C. 16.8 BTU  D. 15.9 BTU Formula: U= mcv T

Description : Find the change in internal energy of 5 lb. of oxygen gas when the temperature changes from 100 ˚F to 120 ˚F. CV = 0.157 BTU/lbm-˚R  A.14.7 BTU  B.15.7 BTU  C. 16.8 BTU  D. 15.9 BTU Formula: U= mcv T

Last Answer : 15.7 BTU

How many Newton’s (N) in 900,000 dynes?  a) 8 Newton’s  b) 9 Newton’s  c) 7 Newton’s  d) 6 Newton’s Formula: 1Newton (N)=100,000dynes

Description : How many Newton’s (N) in 900,000 dynes?  a) 8 Newton’s  b) 9 Newton’s  c) 7 Newton’s  d) 6 Newton’s Formula: 1Newton (N)=100,000dynes

Last Answer : 9 Newton’s

Two masses, one of the 10kg and the other unknown, are placed on a scale in a region where g = 9.67 m/sec2 . The combined weight of these two masses is 174.06 N. Find the unknown mass in kg.  a. 20 kg  b. 19 kg  c. 18 kg  d. 17 kg formula: m=Fg k / g

Description : Two masses, one of the 10kg and the other unknown, are placed on a scale in a region where g = 9.67 m/sec2 . The combined weight of these two masses is 174.06 N. Find the unknown mass in kg.  a. 20 kg  b. 19 kg  c. 18 kg  d. 17 kg formula: m=Fg k / g

Last Answer : 18 kg

A vertical column of water will be supported to what height by standard atmospheric pressure. If the Y w = 62.4lb/ft3 po = 14.7 psi.  a. 44.9 ft  b. 33.9 ft  c. 22.9 ft  d. 55.9 ft formula: ho= po/Yw

Description : A vertical column of water will be supported to what height by standard atmospheric pressure. If the Y w = 62.4lb/ft3 po = 14.7 psi.  a. 44.9 ft  b. 33.9 ft  c. 22.9 ft  d. 55.9 ft formula: ho= po/Yw

Last Answer : 33.9 ft

A mass of 5kg is 100m above a given datum where local g = 9.75 m/s2 . Find the gravitational force in newtons. (Formula: Fg= mg/k )  a. 48.75 N  b. 50 N  c. 45 N  d. None of the above

Description : A mass of 5kg is 100m above a given datum where local g = 9.75 m/s2 . Find the gravitational force in newtons. (Formula: Fg= mg/k )  a. 48.75 N  b. 50 N  c. 45 N  d. None of the above

Last Answer : 48.75 N

A 30-m vertical column of fluid (density 1878 kg/m3 ) is located where g= 9.65 mps2 . Find the pressure at the base of the column. (Formula: pg= gρhg/k )  a. 543680 N/m2  b. 543.68 kPa (gauge)  c. Both a & b  d. None of the above

Description : A 30-m vertical column of fluid (density 1878 kg/m3 ) is located where g= 9.65 mps2 . Find the pressure at the base of the column. (Formula: pg= gρhg/k )  a. 543680 N/m2  b. 543.68 kPa (gauge)  c. Both a & b  d. None of the above

Last Answer : Both a & b

What is the specific weight of water at standard condition? (Formula: γ = ρg / k)  a. 1000 kgm/m3  b. 9.8066 m/s2  c. 1000 kgf/m3  d. None of the above

Description : What is the specific weight of water at standard condition? (Formula: γ = ρg / k)  a. 1000 kgm/m3  b. 9.8066 m/s2  c. 1000 kgf/m3  d. None of the above

Last Answer : 1000 kgf/m3

A system weighing 2kN. Determine the force that accelerate if to 12 m/s^2. a. vertically upward when g = 9.7 m/s^2  A. 4474.23 N  B.5484.23 N  C.4495.23 N  D.5488.23 N Formula: F = m/k (a +g)

Description : A system weighing 2kN. Determine the force that accelerate if to 12 m/s^2. a. vertically upward when g = 9.7 m/s^2  A. 4474.23 N  B.5484.23 N  C.4495.23 N  D.5488.23 N Formula: F = m/k (a +g)

Last Answer : 4474.23 N

3.0 lbm of air are contained at 25 psia and 100 ˚F. Given that Rair = 53.35 ft-lbf/lbm- ˚F, what is the volume of the container?  A.10.7 ft^3  B.14.7 ft^3  C.15 ft^3  D.24.9 ft^3 Formula: use the ideal gas law pV = mRT T = (100 +460) ˚R V = mRT/p

Description : 3.0 lbm of air are contained at 25 psia and 100 ˚F. Given that Rair = 53.35 ft-lbf/lbm- ˚F, what is the volume of the container?  A.10.7 ft^3  B.14.7 ft^3  C.15 ft^3  D.24.9 ft^3 Formula: use the ideal gas law pV = mRT T = (100 +460) ˚R V = mRT/p

Last Answer : 24.9 ft^3

If air is at pressure, p, of 3200 lbf/ft2 , and at a temperature, T, of 800 ˚R, what is the specific volume, v? (R=5303 ft-lbf/lbm-˚R, and air can be modeled as an ideal gas.)  A.9.8 ft^3/lbm  B.11.2 ft^3/lbm  C.13.33 ft^3/lbm  D.14.2 ft^3/lbm Formula: pv = RT v = RT / p

Description : If air is at pressure, p, of 3200 lbf/ft2 , and at a temperature, T, of 800 ˚R, what is the specific volume, v? (R=5303 ft-lbf/lbm-˚R, and air can be modeled as an ideal gas.)  A.9.8 ft^3/lbm  B.11.2 ft^3/lbm  C.13.33 ft^3/lbm  D.14.2 ft^3/lbm Formula: pv = RT v = RT / p

Last Answer : 13.33 ft^3/lbm

The heat absorbed by a unit mass of a material at its holding point in order to convert the material into a gas at the same temperature.  a. Latent Heat of Sublimation  b. Latent Heat of Vaporization  c. Latent Heat of Fusion  d. Latent Heat Of Condensation

Description : The heat absorbed by a unit mass of a material at its holding point in order to convert the material into a gas at the same temperature.  a. Latent Heat of Sublimation  b. Latent Heat of Vaporization  c. Latent Heat of Fusion  d. Latent Heat Of Condensation

Last Answer : Latent Heat of Vaporization

To convert volumetric analysis to gravimetric analysis, the relative volume of each constituent of the flue gases is  (a) divided by its molecular weight  (b) multiplied by its molecular weight  (c) multiplied by its density  (d) multiplied by its specific weight  (e) divided by its specific weight.

Description : To convert volumetric analysis to gravimetric analysis, the relative volume of each constituent of the flue gases is  (a) divided by its molecular weight  (b) multiplied by its molecular weight  (c) ... by its density  (d) multiplied by its specific weight  (e) divided by its specific weight.

Last Answer : Answer : b

A cylinder contains oxygen at a pressure of 10 atm and a temperature of 300 K. The volume of the cylinder is 10 liters. What is the mass of the oxygen in grams? Molecular weight (MW) of oxygen is 32 g/mole?  a. 125.02  b. 130.08  c. 135.05  d. 120.04

Description : A cylinder contains oxygen at a pressure of 10 atm and a temperature of 300 K. The volume of the cylinder is 10 liters. What is the mass of the oxygen in grams? Molecular weight (MW) of oxygen is 32 g/mole?  a. 125.02  b. 130.08  c. 135.05  d. 120.04

Last Answer : 130.08 {(10atm)(10)(32)/(0.0821) (300K)}

Oxygen at 15ºC and 10.3 Mpa gauge pressure occupies 600L. What is the occupied by the oxygen at 8.28 Mpa gauge pressure and 35ºC?  a. 789.32 L  b. 796.32 L  c. 699 L  d. 588 L V2= P1V1/T1P2

Description : Oxygen at 15ºC and 10.3 Mpa gauge pressure occupies 600L. What is the occupied by the oxygen at 8.28 Mpa gauge pressure and 35ºC?  a. 789.32 L  b. 796.32 L  c. 699 L  d. 588 L V2= P1V1/T1P2

Last Answer : 796.32 L

Which of the following is standard temperature and pressure (STP)?  A. 0 degree Celsius and one atmosphere  B. 32 degree Fahrenheit and zero pressure  C. 0 degree Kelvin and one atmosphere  D. 0 degree Fahrenheit and zero pressure

Description : Which of the following is standard temperature and pressure (STP)?  A. 0 degree Celsius and one atmosphere  B. 32 degree Fahrenheit and zero pressure  C. 0 degree Kelvin and one atmosphere  D. 0 degree Fahrenheit and zero pressure

Last Answer : 0 degree Celsius and one atmosphere

The atomic mass of oxygen is  A. 12  B. 14  C. 16  D. 32

Description : The atomic mass of oxygen is  A. 12  B. 14  C. 16  D. 32

Last Answer : Answer: C

The pressure gauge on a 2000 m³ tank of oxygen gas reads 600 kPa. How much volumes will the oxygen occupied at pressure of the outside air 100 kPa?  a) 14026.5 m³  b) 15026.5 m³  c) 13026.5 m³  d) 16026.5 m³ Formula: P1V1/T1 =P2V2/T2

Description : The pressure gauge on a 2000 m³ tank of oxygen gas reads 600 kPa. How much volumes will the oxygen occupied at pressure of the outside air 100 kPa?  a) 14026.5 m³  b) 15026.5 m³  c) 13026.5 m³  d) 16026.5 m³ Formula: P1V1/T1 =P2V2/T2

Last Answer : 14026.5 m³

A 1-kg steam-water mixture at 1.0 MPa is contained in an inflexible tank. Heat is added until the pressure rises to 3.5 MPa and the temperature to 400°. Determine the heat added.  a) 1378.7 kJ  b) 1348.5 kJ  c) 1278,7 kJ  d) 1246,5 kJ Formula: Q = (h2 – p2v2) –(h1 –p1v1)

Description : A 1-kg steam-water mixture at 1.0 MPa is contained in an inflexible tank. Heat is added until the pressure rises to 3.5 MPa and the temperature to 400°. Determine the heat added.  a) 1378.7 kJ  b) 1348.5 kJ  c) 1278,7 kJ  d) 1246,5 kJ Formula: Q = (h2 – p2v2) –(h1 –p1v1)

Last Answer : 1378.7 kJ

A certain gas, with cp = 0.529Btu/ lb. °Rand R = 96.2ft.lb/lb. °R, expands from 5 cu ft and 80°F to 15 cu ft while the pressure remains constant at 15.5psia. Compute for T2.  a.1520°R  b. 1620°R  c. 1720°R  d. 1820°R formula: T2= T1V2/V1

Description : A certain gas, with cp = 0.529Btu/ lb. °Rand R = 96.2ft.lb/lb. °R, expands from 5 cu ft and 80°F to 15 cu ft while the pressure remains constant at 15.5psia. Compute for T2.  a.1520°R  b. 1620°R  c. 1720°R  d. 1820°R formula: T2= T1V2/V1

Last Answer : 1620°R

The flow energy of 5 ft3 of a fluid passing a boundary to a system is 80,000 ft-lb. Determine the pressure at this point.  a. 222 psi  b. 333 psi  c. 444 psi  d. 111 psi formula: Ef= pV

Description : The flow energy of 5 ft3 of a fluid passing a boundary to a system is 80,000 ft-lb. Determine the pressure at this point.  a. 222 psi  b. 333 psi  c. 444 psi  d. 111 psi formula: Ef= pV

Last Answer : 111 psi

A car whose mass is 2 metric tons is accelerated uniformly from stand hill to 100 kmph in 5 sec. Find the driving force in Newton’s.  a. 11,120 N  b. 11,320 N  c. 11,420 N  d. 11520 N formula: F= ma / k

Description : A car whose mass is 2 metric tons is accelerated uniformly from stand hill to 100 kmph in 5 sec. Find the driving force in Newton’s.  a. 11,120 N  b. 11,320 N  c. 11,420 N  d. 11520 N formula: F= ma / k

Last Answer : 11,120N

A certain gas, with cp = 0.529Btu/lb.°R and R = 96.2 ft.lb/lb.°R, expands from 5 cu ft and 80°F to 15 cu ft while the pressure remains constant at 15.5 psia. Compute for T2. (Formula: T2= T1V2/V1)  a. 460°R  b. 270°R  c. 1620 °R  d. None of the above

Description : A certain gas, with cp = 0.529Btu/lb.°R and R = 96.2 ft.lb/lb.°R, expands from 5 cu ft and 80°F to 15 cu ft while the pressure remains constant at 15.5 psia. Compute for T2. (Formula: T2= T1V2/V1)  a. 460°R  b. 270°R  c. 1620 °R  d. None of the above

Last Answer : 1620 °R

There are 1.36 kg of gas, for which R = 377 J/kg.k and k = 1.25, that undergo a nonflow constant volume process from p1 = 551.6 kPa and t1 = 60°C to p2 = 1655 kPa. During the process the gas is internally stirred and there are also added 105.5 kJ of heat. Determine t2. (Formula: T2= T1p2/ p1)  a. 999 K  b. 888 K  c. 456 K  d. One of the above

Description : There are 1.36 kg of gas, for which R = 377 J/kg.k and k = 1.25, that undergo a nonflow constant volume process from p1 = 551.6 kPa and t1 = 60°C to p2 = 1655 kPa. During the process the gas is internally stirred and ... (Formula: T2= T1p2/ p1)  a. 999 K  b. 888 K  c. 456 K  d. One of the above

Last Answer : 999 K

A 10m^3 vessel initially contains 5 m^3 of liquid water and 5 m^3 of saturated water vapor at 100 kPa. Calculate the internal energy of the system using the steam table.  A. 5 x10^5 kJ  B. 8x10^5 kJ  C. 1 x10^6 kJ  D. 2 x10^6 kJ Formula: fromthe steamtable vƒ = 0.001043 m^3 / kg vg = 1.6940 m^3 / kg u ƒ= 417.3 kJ/kg ug= 2506kJ/kg formula: Mvap = V vap/vg M liq = Vliq/ vƒ u =uƒM liq + ug M vap

Description : A 10m^3 vessel initially contains 5 m^3 of liquid water and 5 m^3 of saturated water vapor at 100 kPa. Calculate the internal energy of the system using the steam table.  A. 5 x10^5 kJ  B. 8x10^5 kJ  C. 1 ... 3 kJ/kg ug= 2506kJ/kg formula: Mvap = V vap/vg M liq = Vliq/ vƒ u =uƒM liq + ug M vap

Last Answer : 2 x10^6 kJ

Steam at 1000 lbf/ft^2 pressure and 300˚R has specific volume of 6.5 ft^3/lbm and a specific enthalpy of 9800 lbf-ft/lbm. Find the internal energy per pound mass of steam.  A.2500 lbf-ft/lbm  B.3300 lbf-ft/lbm  C.5400 lbf-ft/lbm  D.6900 lbf-ft/lbm Formula: h= u+ pV u= h– pV

Description : Steam at 1000 lbf/ft^2 pressure and 300˚R has specific volume of 6.5 ft^3/lbm and a specific enthalpy of 9800 lbf-ft/lbm. Find the internal energy per pound mass of steam.  A.2500 lbf-ft/lbm  B.3300 lbf-ft/lbm  C.5400 lbf-ft/lbm  D.6900 lbf-ft/lbm Formula: h= u+ pV u= h– pV

Last Answer : 3300 lbf-ft/lbm

Calculate the power output in horsepower of an 80-kg man that climbs a flight of stairs 3.8 m high in 4.0 s.  a) 744.8 hp  b) 0.998 hp  c) 746 hp  d) 1.998 hp Formula: Power = Fd/t = mgh/t F = W = mg d = h

Description : Calculate the power output in horsepower of an 80-kg man that climbs a flight of stairs 3.8 m high in 4.0 s.  a) 744.8 hp  b) 0.998 hp  c) 746 hp  d) 1.998 hp Formula: Power = Fd/t = mgh/t F = W = mg d = h

Last Answer : 0.998 hp

3 horsepower (hp) = _____________watts?  a) 1492 watts  b) 2238 watts  c) 746 watts  d) 2238 kilowatts Formula: 1hp= 746 watts

Description : 3 horsepower (hp) = _____________watts?  a) 1492 watts  b) 2238 watts  c) 746 watts  d) 2238 kilowatts Formula: 1hp= 746 watts

Last Answer : 2238 watts

Two thick slices of bread, when completely oxidized by the body, can supply 200,000 cal of heat. How much work is this equivalent to?  a) 4,190,000 joules  b) 8,390,000 joules  c) 839,000 joules  d) 419 000 joules Formula: J =Work/Heat J = mechanical equivalent of heat whose value is 4.19 joules/calorie

Description : Two thick slices of bread, when completely oxidized by the body, can supply 200,000 cal of heat. How much work is this equivalent to?  a) 4,190,000 joules  b) 8,390,000 joules  c) 839, ... d) 419 000 joules Formula: J =Work/Heat J = mechanical equivalent of heat whose value is 4.19 joules/calorie

Last Answer : 419 000 joules

How many joules of work is the equivalent of 15000 cal of heat?  a) 62850 joules  b) 3579.95 joules  c) 14995.81 joules  d) 15004.19 joules Formula: J =Work/Heat J = mechanical equivalent of heat whose value is 4.19 joules/calorie

Description : How many joules of work is the equivalent of 15000 cal of heat?  a) 62850 joules  b) 3579.95 joules  c) 14995.81 joules  d) 15004.19 joules Formula: J =Work/Heat J = mechanical equivalent of heat whose value is 4.19 joules/calorie

Last Answer : 62850 joules