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10-1.Thermometry, Thermal Expansion and Calorimetry
medium
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 $g^{-1}{ }^{\circ} C ^{-1}$ )
A
$2.6$
B
$2$
C
$4$
D
$3.2$
(JEE MAIN-2020)
Solution
$\frac{ Cal }{20 gm } \frac{ H _{2} O }{180 gm } \quad \frac{ Steam }{ m }$
$200 \times 1 \times(31-25)$$= m \times 540+ m \times 1 \times(100-31)$
$m \approx 2 \;gm$
Standard 11
Physics
