Quick, Before It Melts

Using blackbody radiation laws, the amount of heat dissipated per second by the ice cube is given by $P_{out} = \sigma A T^4$ , where $A$ is the surface area and $T$ is the temperature of the ice cube in Kelvin. Similarly, the amount of heat absorbed per second by the ice cube is given by $P_{in} = \sigma A T_0^4$ , where $T_0$ is the temperature of the surroundings. Therefore the net heat absorbed by the ice cube per second is $P_{net} = P_{in} - P_{out} = \sigma A (T_0^4 - T^4)$ .

The amount of energy required to melt an ice cube is given by $mL$ , where $m$ is the mass of the ice cube, and $L$ is the specific latent heat of fusion. In the first case, $\dfrac{mL}{t_1} = \sigma A (283^4 - 273^4)$

In the second case, $\dfrac{mL}{t_2} = \sigma A (313^4 - 273^4)$ .

Therefore the time taken in the second case is $\dfrac{283^4 - 273^4}{313^4 - 273^4} \times 100 \approx 21.263$