Question

16 grams of ice at –32°C is to be changed to steam at 172°C. The entire process requires _____ cal. Round your answer to the nearest whole number. The specific heat of both ice and steam is 0.5 cal/g°C. The specific heat of water is 1.00 cal/gK. The heat of fusion is 80 cal/g and the heat of vaporization is 540 cal/g.

Answer 1

Answer:12,352 cal

Explanation:

Given

to change 16 gm ice at $-32^{\circ}C to 172^{\circ}C$

First we need to take ice to $0^{\circ}C$then change its state to water and then raises water temperature from 0 to $100^{\circ}C$ .

After it change water phase to steam and then raise its temperature to $172^{\circ}C$

$heat required(Q)=16\times 0.5\times \left ( 0+32\right )$+$16\times 80+16\times 1\times \left ( 100-0\right )+16\times 540+16\times 0.5\times \left ( 172-100\right )$

Q=12,352 cal