One end of a copper rod of length 1.0 m and area of cross-section 10 − 3 is immersed in boiling water and the other end in ice. If the coefficient of thermal conductivity of copper is 92 c a l / m − s − ∘ C and the latent heat of ice is 8 × 10 4 c a l / k g , then the amount of ice which will melt in one minute is Options: A . &nbsp 9.2 × 10 − 3 k g B . &nbsp 8 × 10 − 3 k g C . &nbsp 6.9 × 10 − 3 k g D . &nbsp 5.4 × 10 − 3 k g

Answer: Option C : C Heat transferred in one minute is utilised in melting the ice so, K A ( θ 1 − θ 2 ) t l = m × L ⇒ m = 10 − 3 × 92 × ( 100 − 0 ) × 60 1 × 8 × 10 4 = 6.9 × 10 − 3 k g

One end of a copper rod of length 1.0 m and area of cross-section 10"

Discussion Forum : Heat Transfer

Question -

One end of a copper rod of length 1.0 m and area of cross-section $10^{- 3}$ is immersed in boiling water and the other end in ice. If the coefficient of thermal conductivity of copper is $92 c a l / m - s -^{\circ} C$ and the latent heat of ice is $8 \times 10^{4} c a l / k g$ , then the amount of ice which will melt in one minute is

Options:

A .

9.2 \times 10^{- 3} k g

B .

8 \times 10^{- 3} k g

C .

6.9 \times 10^{- 3} k g

D .

5.4 \times 10^{- 3} k g

Answer: Option C
:
C

Heat transferred in one minute is utilised in melting the ice so, $\frac{K A (θ_{1} - θ_{2}) t}{l} = m \times L$
$\Rightarrow m = \frac{10^{- 3} \times 92 \times (100 - 0) \times 60}{1 \times 8 \times 10^{4}} = 6.9 \times 10^{- 3} k g$

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