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On a winter day the temperature drops from –15°c to –25°c overnight. if a pan sitting outside contains 0.40 kg of ice, how much heat is removed from the ice for this temperature change?

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taskmasters

the problem can be solve using the formula:
q = m cp dt
where q is the heat removed
cp is the heat capacity of the material ( 2.108 kJ/kg C for ice )
dt is the change temperature

q = (0.40 kg ) ( 2.108 kj/kg C) ( -25 - ( -15) C )
q = -8.432 kj
the negative sign indicates that it is removed

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