Heat and Thermodynamics 1 Question 6
6. Ice at $-20^{\circ} C$ is added to $50 g$ of water at $40^{\circ} C$. When the temperature of the mixture reaches $0^{\circ} C$, it is found that $20 g$ of ice is still unmelted. The amount of ice added to the water was close to (Take, specific heat of water $=4.2 J / g /{ }^{\circ} C$ specific heat of ice $=2.1 J / g /{ }^{\circ} C$ and heat of fusion of water at $\left.0^{\circ} C=334 J / g\right)$
(2019$ Main, 11 Jan I)
(a) $40 g$
(b) $50 g$
(c) $60 g$
(d) $100 g$
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Answer:
Correct Answer: 6. (a)
Solution:
- Let amount of ice be ’ $x$ ’ $g m$.
According to the principle of calorimeter,
heat lost by water $=$ heat gained by ice
Here, heat lost by water
$$ \Delta Q=m s _{\text {water }} \Delta T $$
Substituting the given values, we get
$$ \Delta Q=50 \times 4.2 \times 40 $$
Heat gained by ice,
$$ \begin{aligned} & \Delta Q=x s _{\text {ice }} \Delta T+(x-20) L \\ & =x \times 2.1 \times 20+(x-20) \times 334 \\ & =20 x \times 2.1+334 x-6680 \\ & \therefore 20 x \times 2.1+334 x-6680=50 \times 4.2 \times 40 \\ & 42 x+334 x-6680=8400 \\ & \Rightarrow \quad 376 x=15080 \\ & \text { or } \quad x=40.10 g \\ & x \simeq 40 g \end{aligned} $$
