Given: latent heat of fusion of water 3.33x10^5 J/kg. A 147 g cube of ice at 0 degress C is dropped into 1.3 kg of water that was originally at 79 degrees C. What is the final temperature of the water after the ice melts.

Answer 1

147 g of ice, if it melts entirely, will absorb 0.147*[3.33*10^5 + 4.18 T] Joules
when heating up to temperature T in the liquid state. The water that was originaly liquid will lose heat equal to
1.3*4.18*(79 - T) J.

Set the two equal and solve for T.

Answer 2

Tf=65.99 degress celsius