A beaker contains 200 g of water- The heat capacity of the beaker is equal to that of 20 g of water- The initial temperature of water in the beaker is 20- C- If 440 g of hot water at 92- C is poured in it- the final temperature neglecting radiation loss will be nearest to

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Standard IX

Physics

Specific Heat Capacity

A beaker cont...

Question

A beaker contains $200 g$ of water. The heat capacity of the beaker is equal to that of $20 g$ of water. The initial temperature of water in the beaker is $20^{\circ} C$ . If $440 g$ of hot water at $92^{\circ} C$ is poured in it, the final temperature (neglecting radiation loss) will be nearest to

58^{\circ} C

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68^{\circ} C

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73^{\circ} C

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78^{\circ} C

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Solution

The correct option is B $68^{\circ} C$
The water equivalent of the water+beaker system initially is 220g
Heat released by hot water = heat absorbed by the system
Let final temperature be $t$
Then, $440 \times (92 - t) = 220 (t - 20) \Rightarrow t = 68^{o} C$

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A beaker contains 200g of water. The heat capacity of the beaker is equal to that of 20g of water. The initial temperature of water in the beaker is 20∘C. If 440g of hot water at 92∘C is poured in it, the final temperature (neglecting radiation loss) will be nearest to

The correct option is B 68∘CThe water equivalent of the water+beaker system initially is 220gHeat released by hot water = heat absorbed by the systemLet final temperature be tThen, 440×(92−t)=220(t−20)⇒t=68oC

A beaker contains $200 g$ of water. The heat capacity of the beaker is equal to that of $20 g$ of water. The initial temperature of water in the beaker is $20^{\circ} C$ . If $440 g$ of hot water at $92^{\circ} C$ is poured in it, the final temperature (neglecting radiation loss) will be nearest to

The correct option is B $68^{\circ} C$
The water equivalent of the water+beaker system initially is 220g
Heat released by hot water = heat absorbed by the system
Let final temperature be $t$
Then, $440 \times (92 - t) = 220 (t - 20) \Rightarrow t = 68^{o} C$