posted by Mary .
On a winter day the temperature drops from -5 degree C to -15 degree C overnight. If a pan sitting outside contains 0.40kg of ice, how much heat is removed from the ice for the temperature change? (specific heat (ice) = .5kcal/kg) Is it 02.0kcal?
M*C*delta T = 0.4*0.5*10 = 2.0 kCal
You are right!