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?

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