How much thermal energy is required to change a 40 g ice cube from solid at -10° C to steam at 110° C. [Assume latent heat of fusion for water =80 kcal / kg, specific heat of water =1 k cal / kg ° C, specific heat of Ice =0.5 k cal kg ° C, specific heat of steam =0.48 k cal / kg ° C, latent heat of vaporisation of water- 540 kcal / kg ° C ]

Question

How much thermal energy is required to change a 40 g ice cube from solid at -10° C to steam at 110° C. [Assume latent heat of fusion for water =80 kcal / kg, specific heat of water =1 k cal / kg ° C, specific heat of Ice =0.5 k cal kg ° C, specific heat of steam =0.48 k cal / kg ° C, latent heat of vaporisation of water- 540 kcal / kg ° C ] (A) 29.192 k cal (B) 40.288 k cal (C) 35.188 k cal (D) 30.188 k cal

Tardigrade Admin · Accepted Answer

Following changes occur in given conversion process

So, required thermal energy

= m {c_{ice} (0^{\circ} C - (- 1 0^{\circ} Q) +

L_{fusion} + c_{water} (10 0^{\circ} C - 0^{\circ} C) + L_{vaporisation} + c_{stcam} (11 0^{\circ} C - 10 0^{\circ} C)}

= \frac{400}{1000} {0.5 \times 10 + 80 + 1 \times 100 + 540 + 0.48 \times 10}

= 29.192 k c a l

29.192 k cal

40.288 k cal

35.188 k cal

30.188 k cal

Following changes occur in given conversion process So, required thermal energy =m{c ice ​(0∘C−(−10∘Q)+ Lfusion ​+cwater ​(100∘C−0∘C)+Lvaporisation ​+cstcam ​(110∘C−100∘C)} =1000400​{0.5×10+80+1×100+540+0.48×10} =29.192kcal