Answer:
[tex]680\; \rm s[/tex].
Explanation:
Start by finding the total amount of energy required for melting that much ice.
[tex]\begin{aligned}&\text{Energy required for melting ice sample} \\ &= \text{Mass of Ice} \times \text{Specific Latent Heat of Ice} \\ &= 120\; \rm g \times 340\; \rm J \cdot g^{-1} = 4.08 \times 10^{4}\; \rm J \end{aligned}[/tex].
Hence, the heater would need to supply (at least) [tex]4.08 \times 10^{4}\; \rm J[/tex] of energy.
The power of the heater is [tex]60\; \rm W[/tex], which is equivalent to [tex]60\; \rm J \cdot s^{-1}[/tex], In other words, the heater is rated to supply [tex]60\; \rm J[/tex] of energy every second.
Amount of time it takes for the heater to supply [tex]4.08 \times 10^{4}\; \rm J[/tex] at [tex]60\; \rm J \cdot s^{-1}[/tex]:
[tex]\begin{aligned}\frac{4.08 \times 10^{4}\; \rm J}{60\; \rm J \cdot s^{-1}} = 680\; \rm s\end{aligned}[/tex].
Hence, it would take [tex]680\; \rm s[/tex] for the heater to melt the ice if the heater is insulated, and all the energy from the heater went to the ice.
