One end of a copper rod of length 2 m and area of cross section 10-3 m 2 is immersed in boiling water and the other in solid form of water (ice). If the coefficient of thermal conductivity of copper is 92 cal / m - s - ° C and the latent heat of ice is 9 × 104 cal/kg (assumed) then the amount of ice which will melt in three minute is

Question

One end of a copper rod of length 2 m and area of cross section 10-3 m 2 is immersed in boiling water and the other in solid form of water (ice). If the coefficient of thermal conductivity of copper is 92 cal / m - s - ° C and the latent heat of ice is 9 × 104 cal/kg (assumed) then the amount of ice which will melt in three minute is (A) 15 × 10-1 kg (B) 40 × 10-2 kg (C) 92 × 10-4 kg (D) 25 × 10-2 kg

Tardigrade Admin · Accepted Answer

mL =( KA (Δ T ) t / x ) ⇒ m =(92 × 10-3 × 100 × 180/2 × 9 × 104) m =92 × 10-4 kg

$15 \times 1 0^{- 1} k g$

$40 \times 1 0^{- 2} k g$

$92 \times 1 0^{- 4} k g$

$25 \times 1 0^{- 2} k g$

$m L = \frac{K A ( Δ T ) t}{x}$
$\Rightarrow m = \frac{92 \times 1 0 ^{- 3} \times 100 \times 180}{2 \times 9 \times 1 0 ^{4}}$
$m = 92 \times 1 0^{- 4} k g$

15×10−1kg

40×10−2kg

92×10−4kg

25×10−2kg

mL=xKA(ΔT)t​ ⇒m=2×9×10492×10−3×100×180​ m=92×10−4kg

$15 \times 1 0^{- 1} k g$

$40 \times 1 0^{- 2} k g$

$92 \times 1 0^{- 4} k g$

$25 \times 1 0^{- 2} k g$

$m L = \frac{K A ( Δ T ) t}{x}$
$\Rightarrow m = \frac{92 \times 1 0 ^{- 3} \times 100 \times 180}{2 \times 9 \times 1 0 ^{4}}$
$m = 92 \times 1 0^{- 4} k g$