Question

Water is boiled in a container having a bottom of surface area 25 cm², thickness 1.0 mm and thermal conductivity 50 W m⁻¹°C⁻¹. 100 g of water is converted into steam per minute in the steady state after the boiling starts. Assuming that no heat is lost to the atmosphere, calculate the temperature of the lower surface of the bottom. Latent heat of vaporisation of water = 2.26 × 10⁶ J kg⁻¹.

Answer 1

Area of the bottom of the container, $A=25{\mathrm{cm}}^2=25\times {10}^{-4}{\mathrm{m}}^2$
Thickness of the bottom of the container, l = 1 mm = 10^$-$3 m
Latent heat of vaporisation of water, $L=2.26\times {10}^6\mathrm{J}{\mathrm{kg}}^{-1}$
Thermal conductivity of the container, $k=50\mathrm{W}{\mathrm{m}}^{-1}°{\mathrm{C}}^{-1}$
mass = 100 g = 0.1 kg

Rate of heat transfer from the base of the container is given by
$$\begin{gathered} \frac{\Delta Q}{\Delta t}=\frac{mL}{\Delta t}=\frac{\left(0.1\right)\times 2.26\times {10}^6}{1\min } \\ \frac{\Delta Q}{\Delta t}=0.376\times {10}^4\mathrm{J}/\mathrm{s} \end{gathered}$$
Also,
$$\begin{gathered} \frac{\Delta Q}{\Delta t}=\frac{\Delta T}{{\displaystyle \frac{l}{kA}}} \\ \Rightarrow 0.376\times {10}^4=\frac{T-100}{{\displaystyle \frac{{10}^{-3}}{50\times 25\times 10{}^{-4}}}} \\ \Rightarrow 0.376\times {10}^4=\frac{50\times 25\times {10}^{-4}\left(T-100\right)}{{10}^{-3}} \\ \Rightarrow \left(T-100\right)=3.008\times 10 \\ \Rightarrow T=130°\mathrm{C} \end{gathered}$$