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Question: Work done in converting one gram of ice at – 10<sup>0</sup>*C* into steam at 100<sup>0</sup>*C* is...

Work done in converting one gram of ice at – 100C into steam at 1000C is

A

3045 J

B

6056 J

C

721 J

D

616 J

Answer

3045 J

Explanation

Solution

Work done in converting 1gm of ice at – 100C to steam at 1000C

= Heat supplied to raise temperature of 1gm of ice from – 100C to 00C [m × cice × ∆T]

+ Heat supplied to convert 1 gm ice into water at 00C [m × Lice]

+ Heat supplied to raise temperature of 1gm of water from 00C to 1000C [m × cwater × ∆T]

+ Heat supplied to convert 1 gm water into steam at 1000C [m × Lvapour]

= [m × cice × ∆T] + [m × Lice] + [m × cwater × ∆T] + [m × Lvapour]

= [1×0.5×10]+[1×80]+[1×1×100]+[1×540]\lbrack 1 \times 0.5 \times 10\rbrack + \lbrack 1 \times 80\rbrack + \lbrack 1 \times 1 \times 100\rbrack + \lbrack 1 \times 540\rbrack = 725calorie=725×4.2=3045J725calorie = 725 \times 4.2 = 3045J