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Standard X

Physics

Specific Heat Capacity

A refrigerato...

Question

A refrigerator converts $100 g$ of water at $20^{o} C$ to ice at - $10^{o} C$ in $73.5 m i n$ . Calculate the average rate of heat extraction in Watt. The specific heat capacity of water is $4.2 J g^{- 1} K^{- 1}$ , specific latent heat of ice is $336 J g^{- 1}$ and the specific heat capacity of ice is $2.1 J g^{- 1} K^{- 1}$

5 W

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10 W

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20 W

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100 W

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Solution

The correct option is B $10 W$
$m = 0.1 k g$
heat extracted $Q = 0.1 \times 4200 (20 - 0) + 0.1 \times 336000 + 0.1 \times 2100 \times (0 - (- 10))$
$t = 73.5 m i n o r 4530 s e c$
average rate of heat extraction = $\frac{Q}{t} = 10 W$

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Similar questions

Q. A refrigirator converts

100 g

of water at

20^{o} C

to ice at

- 10^{o} C

35

min. Calculate the average rate of heat extraction in watt. The specific heat capacity of water is

4.2 J g^{- 1}

K^{- 1}

, specific heat of ice is

336 J g^{- 1}

and the specific heat capacity of ice is

2.1 J g^{- 1}

K^{- 1}

Q. Calculate the total amount of heat energy required to convert

100 g

of ice at

- 10^{o} C

completely into water at

100^{o} C

. Specific heat capacity of ice

= 2.1 J g^{- 1} K^{- 1}

, specific heat capacity of water

= 4.2 J g^{- 1} K^{- 1}

, specific latent heat of ice

= 336 J g^{- 1}

Q. A refrigerator converts

100 g

of water at

20^{o} C

to ice at

- 10^{o} C

35

minutes. Calculate the average rate of heat extraction in terms of watts.
Given:
Specific heat capacity of ice

= 2.1 J g^{- 1}^{o} C^{- 1}

Specific heat capacity of water

= 4.2 J g^{- 1}^{o} C^{- 1}

Specific Latent heat of fusion of ice

336 J

g^{- 1}

Q. Find the result of mixing

10 g

of ice at

- 10^{o} C

with

10 g

of water at

10^{o} C

. Specific heat capacity of ice

= 2.1 J g^{- 1} K^{- 1}

, specific latent heat of ice

= 336 J g^{- 1}

and specific heat capacity of water

= 4.2 J g^{- 1} K^{- 1}

Q. How much heat energy is released when

5.0 g

of water at

20^{o} C

changes into ice at

0^{o} C

? Take specific heat capacity of water

= 4.2 J g^{- 1} K^{- 1}

, specific latent heat of fusion of ice

= 336 J g^{- 1}

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A refrigerator converts 100 g of water at 20oC to ice at - 10oC in 73.5 min. Calculate the average rate of heat extraction in Watt. The specific heat capacity of water is 4.2Jg−1K−1, specific latent heat of ice is 336Jg−1 and the specific heat capacity of ice is 2.1Jg−1K−1

The correct option is B 10 Wm=0.1 kgheat extracted Q=0.1×4200 (20−0)+0.1×336000+0.1×2100×(0−(−10)) t=73.5 min or 4530 secaverage rate of heat extraction =Qt=10 W

The correct option is B $10 W$
$m = 0.1 k g$
heat extracted $Q = 0.1 \times 4200 (20 - 0) + 0.1 \times 336000 + 0.1 \times 2100 \times (0 - (- 10))$
$t = 73.5 m i n o r 4530 s e c$
average rate of heat extraction = $\frac{Q}{t} = 10 W$