The only possibility of heat flow in a thermos flask is through its cork which is 75 cm 2 in area and 5 cm thick- Its thermal conductivity is 0-0075 cal - cm sec - C- The outside temperature is 40- C and latent heat of ice is 80 cal g -1- Time taken by 500 g of ice at 0- C in the flask to melt into water at 0- C isA- 2-47 hrB- 4-27 hrC- 7-42 hrD- 4-72 hr

The correct option is A 2.47 hr mL = KA θ t/ x⇒ 500 × 80 = 0.0075 × 75 ×(40 0)t/5 ⇒ t = 8.9 × 10^3 sec = 2.47 hr. ...

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Standard X

Physics

Specific Heat Capacity

The only poss...

Question

The only possibility of heat flow in a thermos flask is through its cork which is 75 $c m^{2}$ in area and 5 cm thick. Its thermal conductivity is 0.0075 cal/cm $s e c^{\circ} C$ . The outside temperature is $40^{\circ} C$ and latent heat of ice is 80 cal $g^{- 1}$ . Time taken by 500 g of ice at $0^{\circ} C$ in the flask to melt into water at $0^{\circ} C$ is

2.47 hr

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4.27 hr

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7.42 hr

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4.72 hr

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Solution

The correct option is A
2.47 hr

mL = $\frac{K A △ θ t}{△ x} \Rightarrow 500 \times 80 = \frac{0.0075 \times 75 \times (40 - 0) t}{5}$

$\Rightarrow t = 8.9 \times 10^{3} s e c = 2.47 h r .$

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The correct option is A 2.47 hr mL = KA △ θ t△ x⇒500 × 80 = 0.0075 × 75 ×(40 − 0)t5 ⇒t = 8.9 × 103 sec = 2.47 hr.

The correct option is A
2.47 hr

mL = $\frac{K A △ θ t}{△ x} \Rightarrow 500 \times 80 = \frac{0.0075 \times 75 \times (40 - 0) t}{5}$

$\Rightarrow t = 8.9 \times 10^{3} s e c = 2.47 h r .$