Question #4907

A big steam boiler is made of steel and weighs 900 kilograms empty. When filled with water, the boiler can hold 400 kilograms of water. Assuming that only 70% of the heat is delivered to the boiler steel and water (the rest goes up the chimminey) How much heat is required to raise the temperature of the entire apparatus from 20 degrees C to 100 degrees C? The specific heat of steel is 0.46 Joules/gram-degree C. <br>a. 238 megaJoules <br>b. 374 megaJoules <br>c. 418 megaJoules <br>d. 593 megaJoules <br>e. 619 megaJoules

Expert's answer

n(Energy conversion efficiency) = 0.7 or 70%

m(steel) = 900 kg = 900 000

g

m(water) = 400 kg = 400 000 g

s(steel) = 0.46 J/(g*C)

s(water) =

4.186 J/(g*C)

t1 = 20 C

t2 = 100 C

dT = 100 - 20 = 80 C

Q(100%) =

Q(steel) + Q(water)

Q(100%) = m(steel) * s(steel) * dT + m(water) * s(water)

* dT

Q(100%) = 900 000 g * 0.46 J/(g*C) * 80 C + 400 000 g * 4.186 J/(g*C) *

80 C = 167.1 MJ

Q(70%) = Q(100%) / 0.7 = 238.7 MJ

The right answer is a.

238 megaJoules.

m(steel) = 900 kg = 900 000

g

m(water) = 400 kg = 400 000 g

s(steel) = 0.46 J/(g*C)

s(water) =

4.186 J/(g*C)

t1 = 20 C

t2 = 100 C

dT = 100 - 20 = 80 C

Q(100%) =

Q(steel) + Q(water)

Q(100%) = m(steel) * s(steel) * dT + m(water) * s(water)

* dT

Q(100%) = 900 000 g * 0.46 J/(g*C) * 80 C + 400 000 g * 4.186 J/(g*C) *

80 C = 167.1 MJ

Q(70%) = Q(100%) / 0.7 = 238.7 MJ

The right answer is a.

238 megaJoules.

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