3. #1 points 12 P075.Tutorial. My Notes Ask Your Teacher A 0.250-kg aluminum bow

Question

3. #1 points 12 P075.Tutorial. My Notes Ask Your Teacher A 0.250-kg aluminum bowl holding 0.800 kg of soup at 25.8°C is placed in a freezer, what is the final temperature if 422 kJ of energy is transferred from the bowl and soup? Assume the soup has the same thermal properties as that of water, the specific heat of the liquid soup is 1.00 kcal/(kg·°C), frozen soup is 0.500 kcal/(kg·°C), and the latent heat of fusion is 79.8 kcal/kg. The specific heat of aluminum is 0.215 kcal/(kg·°C). oC Tutorial

Explanation / Answer

Let final temperature be t°C below freezing point

Total heat transferred = heat given out by al bowl + heat given out by liquid + heat given out when liquid freezes + heat given out by solid ( ice)

422KJ= 100.86 K cal

100.86= .250*.215*(t+25.8) + .800*1*25.8 +.800*79.8 +.800*.5*t

14.993= 0.45375t

T = -33.04°c

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