1 ,kg of ice at -20^{circ} C is mixed with 2 ,kg of water at 90^{circ} C . Assuming that there is no

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10-1.Thermometry, Thermal Expansion and Calorimetry

medium

$1 \,kg$ of ice at $-20^{\circ} C$ is mixed with $2 \,kg$ of water at $90^{\circ} C$. Assuming that there is no loss of energy to the environment, the final temperature of the mixture is ............ $^{\circ} C$ (Assume, latent heat of ice $=334.4 \,kJ / kg$, specific heat of water and ice are $4.18 \,kJ kg ^{-1} K ^{-1}$ and $2.09 \,kJ kg ^{-1}- K ^{-1}$, respectively.)

$30$

$0$

$80$

$45$

(KVPY-2015)

Solution

(a)

Let final temperature of mixture is $T^{\circ} C$. Then,

Heat lost by $2 \,kg$ water at $90^{\circ} C$ to cool down at $T^{\circ} C =$ Heat gained by $1 \,kg$ ice at $-20^{\circ} C$ to reach at $0^{\circ} C +$ Heat gained by $1 \,kg$ ice at $0^{\circ} C$ to change its state from ice to water + Water $1 \,kg$ formed at $0^{\circ} C$ is now absorbs heat to reach temperature of $T^{\circ} C$

$\Rightarrow m_w \varepsilon_w \Delta T=m_i \varepsilon_i\left(0-\left(-20^{\circ} C \right)\right)$ $+m_i L+m_i s_w(T-0)$

$\Rightarrow 2 \times 418 \times(90-T)=1 \times 2.09 \times 20$ $+1 \times 334.4+1 \times 418 \times T$

$\Rightarrow 7524-376.2=3 \times 418 \times T$

$\Rightarrow T=30^{\circ} C$

So, final temperature of mixture is $30^{\circ} C$.

Standard 11

Physics