A 5.0 kg block of steel with an initial temperature of 90°C is placed in a styro

Question

A 5.0 kg block of steel with an initial temperature of 90°C is placed in a styrofoam cup containing 0.05 kg of ice at -10°C. Assuming that no heat escapes from the cup, what is the final temperature of the block? The specific heat of steel is 450 J/kg*K, specific heat of ice is 2090 J/kg*K, the latent heat of fusion of water is 3.33 X 105 J/kg, and the specific heat of water is 4186 J/kg*K

Options:

a.) Tfinal steel = -5.9°C

b.) Tfinal steel = 5.9°C

c.) Tfinal steel = 8.3°C

d.) Tfinal steel = 44.2°C

Explanation / Answer

heat loss = heat gain

heat loss by steel block = ms*cs*change in temperature

Qloss = 5*450*(363-T).....(1)

heat gain = 0.05*2090(0-(-263))+0.05*3.33*10^5+0.05*4186*(T-0).....(2)

from equations 1 and 2

816750-2250T = 27483.5+16500+209.3T

T = 314.2 K

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