Question

An electric heater converts ice at 0^oC in a bucket into water in 10 minutes. The same heater heats the same water to 100^oC in another 12.5 minutes. The latent heat of ice is(neglect heat capacity of bucket)(Specific heat of water is 1 cal/g^oC)?

Answer 1

Dear Student ,
Here in this case ,
$$\begin{gathered} Q\times t_1=mLfortheicetobecomewaterand \\ Q\times t_2=ms∆tforthewatertobecomemorehotter. \\ So,here, \\ \frac{mL}{10\times 60}=\frac{ms∆t}{12.5\times 60} \\ \Rightarrow L=\frac{4}{5}\times 1\times 100=80cal/g \end{gathered}$$
Regards