Question

Calculate the amount of heat reqiured to convert $1\text{kg}$ of ice at $-{10}^{\circ }\text{C}$ into steam at ${100}^{\circ }\text{C}$. Specific heat capacity of ice is$=2100\text{J}{kg}^{-1}{K}^{-1}$, latent heat of fusion of ice $=3.36\times {10}^5\text{J}{kg}^{-1}$, specific heat capacity of water $=4200\text{J}{kg}^{-1}{K}^{-1}$ and latent heat of vapourization of water$=2.25\times {10}^6\text{J}{kg}^{-1}$.

• A. ${10}^6\text{J}$
• B. $2\times {10}^6\text{J}$
• C. $3.03\times {10}^6\text{J}$
• D. $4.5\times {10}^6\text{J}$

Answer 1

The correct option is C $3.03\times {10}^6\text{J}$

Heat required to take the ice from $-{10}^{\circ }\text{C}$ to $0^{\circ }\text{C}$:
$Q_1=m{s}_{ice}\Delta T=1\times 2100\times 10=21000\text{J}$

Heat required to melt the ice at $0^{\circ }\text{C}$ to water:
$Q_2=m{L}_{fusion}=1\times 3.36\times {10}^5=336000\text{J}$

Heat required to take $1\text{kg}$ of water from $0^{\circ }\text{C}$ to ${100}^{\circ }\text{C}$:
$Q_3=m{s}_{water}\Delta T=1\times 4200\times 100=420000\text{J}$

Heat required to convert $1\text{kg}$ of water at ${100}^{\circ }\text{C}$ into steam:
$Q_4=m{L}_{vapour}=1\times 2.25\times {10}^6=2250000\text{J}$

Total heat required in the entire process is given by:
$Q_1+Q_2+Q_3+Q_4=21000+336000+420000+2250000=3.03\times {10}^6\text{J}$.