Question

Ice of mass 200 g at $0^{\circ }C$ is converted to water at $0^{\circ }C$ in 8 minutes supplying heat to it at a constant rate. Calculate the time taken to raise the temperature of water to ${100}^{\circ }C$ with heat provided at the same rate. (take the latent heat of fusion of ice $=336J{g}^{-1}{}^{\circ }{C}^{-1}$ and specific heat capacity of water $=4.2J{g}^{-1}{}^{\circ }{C}^{-1}$)

Answer 1

Given: Mass of ice = $200gm$

initial temperature of ice = $0°C$

Latent heat of fusion of ice $=336J{g}^{-1}{}^o{C}^{-1}$

Specific heat capacity of water $=4.2J{g}^{-1}°C^{-1}$

Heat required to convert to ice into water $=Q_1$

$Q_1=mL(m=mass,L=Latentheat)=200\times 336=67200Joules$

Now it is given that heat is supplied as constant rate:

In 8 minutes heat required = $67200Joules$

In 1 second heat required $=\frac{67200}{8\times 60}Joules=\frac{672}{48}\times 10=140Joules$

Constant rate = $140J/s$

Now heat required to change the temperature of water from $0°C$ to $100°C=Q_2$

$Q_2=mc\Delta T=200\times 4.2\times 100=84000J$

(c=specific heat capacity of water)

140 Joules is supplied in = $1s$

1 Joule is supplied in $=\frac{1}{140}second$

84000 Joules is supplied $=\frac{1}{140}\times 84000second=600second=10minutes$