Question

Equal amounts of ice at $0^{0}C$ are placed in three containers P, Q and R, which are kept in a constant temperature environment. Identical heating elements are placed inside each container. These heating elements are powered with different voltages:100 V, 200 V and 300 V in containers P, Q and R, respectively.It was found that it took 20 minutes to melt all the ice in container Q, and it took 4 minutes to melt all the ice in container R. Assuming that at any instant heat is uniformly dissipated in each container throughout its volume, which of the following is correct?

• A. It will take (approximately) 80 min to melt all the ice in container P.
• B. It will take (approximately) 100 min to melt all the ice in container P.
• C. It will take (approximately) 132 min to melt all the ice in container P.
• D. It will not be possible to melt all the ice in container P with the given power source.

Answer 1

Answer: (D)

It will not be possible to melt all the ice in container P with the given power source.

$V_1=300V$$t_1=5min$
$V_2=200V$
$t_2=20min$$V_3=100V$$t_3=?min$$Rateofheatsuppliedbyheatingcontainer=\frac{V^2}{R}$
For first and second container (resistance ${}^R$ is same)
$\frac{t_1}{t_2}=\frac{1}{4}\ne \frac{V_1^2}{V_2^2}=\frac{4}{9}$
Second container should take 2.25 times that of container one but it is taking four times that of container one. There is some heat loss ${(}_{loss}^P)$ to the environment which is at lower temperature than ice temperature $\left(0^{0}C\right)$ .
Same amount of heat is being melt in both containers, that means amount of heat supplied is same.
${(\frac{V_1^2}{R}-P_{loss})}_1={(\frac{V_2^2}{R}-P_{loss})}_2$
which gives $P_{loss}R=\frac{5}{3}\times {10}^4=V_{loss}^2>V_3^2={10}^4$
Container will loose all heat given by 100 V power supply and ice will cool down instead of melting.
Additional:
Say outside temp is ${}^{T}0$ and final temperature of ice is ${}^{T}f$
.$k(0-T_0)={10}^4\times \frac{5}{3}$
$k(T_f-T_0)={10}^4\times 1$
which gives $T_f=\frac{2}{3}{T}_0$