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

Step-by-step 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* 0.5* \left ( 0+32\right )`+
`16* 80+16* 1* \left ( 100-0\right )+16* 540+16* 0.5* \left ( 172-100\right )`

Q=12,352 cal