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
