A copper rod has a length of 1.7 m and a cross-sectional

Ответ:

shelovejaylocs

14.02.2022

Given :

A copper rod has a length of 1.7 m and a cross-sectional area of $3.2 \times 10^{-4} \ m^2$ .

One end of the rod is in contact with boiling water and the other with a mixture of ice and water.

To Find :

The mass of ice per second that melts.

Solution :

Amount of heat transfer from boiling water to ice through the copper bar is :

$Q = \dfrac{kA\Delta T }{L}\\\\Q = \dfrac{390\times 3.2 \times 10^{-4} \times 100}{1.7}\\\\Q = 7.34 \ J/s$

Let, amount of ice melts per second is :

$m=\dfrac{Q}{L_f}$

Here, L_f is the latent of fusion of water and is equal to $L_f = 3.34\times 10^5 \ J/s$ .

Putting all values in above equation, we get :

$m=\dfrac{7.34}{3.34\times 10^5}\\\\m = 2.19 \times 10^{-5} \ kg/s$

Hence, this is the required solution.

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