Water\'s heat of fusion is 80. cal/g , and its specific heat is 1.0calgC . Some

Question

Water's heat of fusion is 80. cal/g , and its specific heat is 1.0calgC .

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?

Water's heat of fusion is 80. cal/g , its specific heat is 1.0calgC , and its heat of vaporization is 540 cal/g .

A canister is filled with 390 g of ice and 100. g of liquid water, both at 0 C . The canister is placed in an oven until all the H2O has boiled off and the canister is empty. How much energy in calories was absorbed?

Explanation / Answer

1. Water losses heat of fusion at 0 deg.c to become liquid at 0 deg.c. This energy is = Heat of fusion (Cal/g)* mass fo water= 80*1400 Cal =112000 cal

2. liquid at 0deg.c gains sensbile heat to become liquid at 32 deg.c and sensible heat is given by= mass* specific heat* temperature difference= 1400*1*32=44800 cal

total heat energy= sum of heat of fusion + sensible heat= 112000+44800 =156800 Cal

II

a). For 390gm of ice gains sensible heat to become liquid at 0 deg.c and this heat energy is= heat of fusion of ice* latent heat of fusion of ice= 390*80=31200 cal

b) now all the mass of 390 g of ice and 100 gm of liquid water at 0 deg.c, total mass= 390+100=490gm. Sensible heat of 490 gms of water to take water from 0 deg.c to 100 deg.c= mass*specific heat* temperature difference= 490*1*100=49000 cal

c) at 100 deg.c, water needs latent heat of fusion to become vapor and this heat= 490*540 = 264600 cal

Total heat to be supplied= sum of a+b+C= 31200+49000+264600 Cal=344800 Cal

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