Question

< previous 8 of 8 to Submit My Answers Give Up Part B cal Water's heat of fusion is 80. cal/g , and its specific heat is 1.0 Some velomobile seats have been designed to hold ice packs inside their cushions. If you started a nde with ice packs that held 1200 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? Express your answer to two significant figures and include the appropriate units. Hints Value Units Submit My Answers Give Up

Answer 1

ANSWER--

conversion of frozen ice at 0 degree C to water at 0 degree C.

Q1= m x L , where m= mass of water ; L = heat of fusion

Q1 = 1200 X 80 = 96000 cal

Now the conversion of water at 0 degree C to water at 37 degree C

Q2= m X c X $\Delta T$

Q2 = 1200X 1.0 X (32 - 0)

Q2= 38400 cal

Therefore the total heat absorbed :-

Q= Q1+Q2

Q= 96000+38400 cal

Q= 134,400 cal

Answer ~ total energy absorbed= 134,400 cal