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 - cmsec 0 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=KAθ 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 XII

Physics

Thermal Resistance

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 $c a l / c m s e c^{o} C$ . The outside temperature is $40^{o} C$ and latent heat of ice is 80 $c a l g^{- 1}$ . Time taken by 500 g of ice at $0^{o} C$ in the flask to melt into water at $0^{o} C$ 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

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

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The correct option is A 2.47 hr mL=KA△θt△x⇒500×80=0.0075×75×(40−0)t5

The correct option is A
2.47 hr

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