Question

The amount of heat required to convert $1g$ of ice at $-{10}^o$C into steam at ${100}^o$C is (specific heat of ice $=0.5$ cal/g/${}^o$C, latent heat of ice $=80$ cal/g, latent heat of steam $=540$ cal/g)

• A. $3031$J
• B. $735$J
• C. $1000$J
• D. $4200$J

Answer 1

The correct option is A $3031$J

$Q=M\times S_1(T_2-T_1)+M\times L_1+M\times S_2\times (T_3-T_2)+M\times L_2$ ,Where

$L_1=$ latent heat of fusion of water $334J/g$ .

$L_2=$ latent heat of vaporisation of water $2258J/g$

$S_1=$ specific heat of ice $0.50cal/g=2.093J/g$

$S_2=$ specific heat of water $4.186J/g$

$Q=1\times 2.093\times (0+10)+1\times 334+1\times 4.186\times (100-0)+1\times 2258$

$Q=3031.53J$

Or $Q=724.20calorie$