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  4. A calorimeter of water equivalent 20 g contains 180 g of water at 25° C. 'm' grams of steam at 100° C is mixed in it till the temperature of the mixure is 31° C. 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)

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

JEE MainJEE Main 2020Thermal Properties of Matter

Solution:

$\frac{ Cal }{20\, gm } \frac{ H _{2} O }{180 gm } \quad \frac{\text { Sterm }}{ m }$
$25^{\circ} C \quad 25^{\circ} C \quad 100^{\circ} C$
$200 \times 1 \times(31-25)$
$= m \times 540+ m \times 1 \times(100-31)$