Question

1g of steam is sent into 1g of ice. The resultant temperature of the mixture is:

(Latent heat of ice $L_{ice}=334J$ , $c=4.2J$, latent heat of vaporization $=2258J$)

• A. ${270}^{0}C$
• B. ${230}^{0}C$
• C. ${100}^{0}C$
• D. ${50}^{0}C$

Answer 1

The correct option is C ${100}^{0}C$

Energy required to convert 1g of ice at 0${}^o$C to 100${}^o$C water is

$=m{L}_{ice}+mc\Delta T=1[334+4.2(100-0\left)\right]=754J$

Latent heat of vaporization of 1g steam$=$$2260J$

So, when a gram of steam is mixed with one gram of ice, ice converts to water at 100${}^o$C and a part of steam converts to water at 100${}^o$C.

The resultant mixture will be at 100${}^o$C.