Question

Work done in converting one gram of ice at $–{10}^{\circ }C$ into steam at ${100}^{\circ }C$ is

• A. 3045 J
• B. 6056 J
• C. 721 J
• D. 616 J

Answer 1

The correct option is A 3045 J

Work done in converting 1gm of ice at $–{10}^{\circ }C$ to steam at ${100}^{\circ }C$
= Heat supplied to raise temperature of 1gm of ice from $-{10}^{\circ }Cto{0}^{\circ }C[m\times C_{ice}\times \Delta T]$
+ Heat supplied to convert 1 gm ice into water at $0^{\circ }C[m\times L_{ice}]$
+ Heat supplied to raise temperature of 1gm of water from
$0^{\circ }Cto{100}^{\circ }C[m\times C_{water}\times \Delta T]$
+ Heat supplied to convert 1 gm water into steam at ${100}^{\circ }C[m\times L_{vapour}]$
$=[m\times C_{ice}\times \Delta T]+[m\times L_{ice}]+[m\times C_{water}\times \Delta T]+[m\times L_{vapour}]=[1\times 0.5\times 10]+[1\times 80]+[1\times 1\times 100]+[1\times 540]=725calorie=725\times 4.2=3045J$