Answered

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 :

Answer:12,352 cal

Explanation:

Given

to change 16 gm ice at [tex]-32^{\circ}C to 172^{\circ}C[/tex]

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

After it change water phase to steam and then raise its temperature to [tex]172^{\circ}C[/tex]

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

Q=12,352 cal

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