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Q.

A calorimeter of water equivalent 20g contains 180g of water at $25^{0} C$ . ‘m’ grams of steam at $100^{0} C$ is mixed in it till the temperature of the mixture is $31^{0} C$ . The value of ‘m’ is close to (Latent heat of water=540 cal $g^{- 1}$ ,specific heat of water=1 cal $g^{- 1} {}^{0}C^{- 1}$ )

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a

2

b

4

c

3.2

d

2.6

answer is A.

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Detailed Solution

Water equivalent (w)=20g

Amount of water =180g at $25^{0} C$

m gram steam at $100^{0} C$

Final temperature (T)= $31^{0} C$

Heat lost by steam =Heat given by (water +calorimeter)

$m \times L + m S_{w} Δ T_{1}$ $= (m_{ω} S_{ω} + m_{c} S_{C}) Δ T_{2}$

$\Rightarrow m \times 540 + m \times 1 (100 - 31)$ $= (180 \times 1 + 20 \times 1) \times (31 - 25)$

$540 m + 69 m = 200 \times 6$

So, $m = \frac{1200}{609} = 1.97$ = 2g

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A calorimeter of water equivalent 20g contains 180g of water at 250C. ‘m’ grams of steam at 1000C is mixed in it till the temperature of the mixture is 310C. The value of ‘m’ is close to (Latent heat of water=540 cal g−1,specific heat of water=1 cal g−1 C −1 0)