anonymous one year ago Water's heat of fusion is 80. cal/g , and its specific heat is 1.0calg⋅∘C .Some velomobile seats have been designed to hold ice packs inside their cushions. If you started a ride with ice packs that held 1400 g of frozen water at 0 ∘C , and the temperature of the water at the end of the ride was 32 ∘C , how many calories of heat energy were absorbed?

the requested amount $$\large Q_T$$ of heat is given by the heat needed to melting the ice, which is: $\Large {Q_1} = 80 \cdot 1400 = ...cal$ plus the heat needde to raise the temperature of the resulatant water at 0°C to 32°C, which is: $\Large {Q_2} = 1 \cdot 1400 \cdot \left( {32 - 0} \right) = ...cal$ namely: $\Large {Q_T} = {Q_1} + {Q_2} = ...cal$