Question

if you supply 3600 kJ of heat, how many grams of ice at 0°C can be melted, heated to its boiling point? (Make M your X in equations) (figure out which q you need)

Answer 1

=1.36kg

Step-by-step explanation:

First heat is absorbed to melt the ice- latent heat of fusion of ice then heat is absorbed to raise the temperature to 373K

The specific heat capacity of water is 4.187kJ/kg/K while the latent heat of fusion of ice is 2230kJ/kg/K.

Letting the mass of the ice to be =x

x×2230kJ/kg/K + x×4.187kJ/kg/K×100K= 3600kJ

2230x+418.7x=3600

2647.8x=3600

x=1.36 kg