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 1900g 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?

Question

aaronq · Answer

Given: 1900g of frozen water at 0 ∘C  (1900 g of ice)

            $\Delta H_{fusion}=80\;cal/g$ ; $C_p^{\;water}=1\;cal/g*^oC$

Find the heat req'd to melt the ice: $q_{fusion}=\Delta H_{fusion}*m_{ice}$

Now, find the heat req'd to raise the temp of the water to -32 centigrade
             
          $q_{water}=m_{water}*C_p^{\;water}*\Delta T$;          where, $\Delta T=T_f-T_i$

Add the heat of both to get the total heat absorbed:   $q_{total}=q_{fusion}+q_{water}$

aaronq · Answer

i meant 32 not "-32"