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System Ice
Frequently Asked Questions...
How much energy must be removed from this system by the refrigerator to turn the water into ice at -7.0°C?
A volume of 0.60 L of water at 16°C is put into an aluminum ice-cube tray of mass 0.310 kg at the same temperature. How much energy must be removed from this system by the refrigerator to turn the water into ice at -7.0°C?
Answer:
Q required = Q to cool water to 0oC + Q to freeze water + Q to cool ice to -7oC + Q to cool aluminum to -7oC = m*c*deltaT + m*Lf + m*c*deltaT)ice + m*c*deltaT Al
So Q = 0.60*4186*16 + 0.60*3.35x10^5 + 0.60*2090*7 + 0.310*910*23 = 2.56x10^5J
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