Question 165: Thermodynamics & KTG

Question

10 kg of ice at $-10^{\circ} \mathrm{C}$ is added to 100 kg of water to lower its temperature from 25 ${ }^{\circ} \mathrm{C}$. Consider no heat exchange to surroundings. The decrement to the temperature of water is $\_\_\_\_$ ${ }^{\circ} \mathrm{C}$.
(specific heat of ice $=2100 \mathrm{~J} / \mathrm{Kg} .{ }^{\circ} \mathrm{C}$, specific heat of water $=4200 \mathrm{~J} / \mathrm{Kg} .{ }^{\circ} \mathrm{C}$, latent heat of fusion of ice $=3.36 	imes 10^5 \mathrm{~J} / \mathrm{Kg}$ )

Accepted Answer

Mass of the ice is $\mathrm{m}_{\mathrm{i}}=10 \mathrm{~kg}$,
Initial temperature of ice is $\mathrm{T}_{\mathrm{i}}=-10^{\circ} \mathrm{C}$,
The specific heat of ice is $\mathrm{c}_{\mathrm{i}}=2100 \frac{\mathrm{~J}}{\mathrm{~kg} \cdot{ }^{\circ} \mathrm{C}}$,
The latent heat of fusion of ice is $\mathrm{L}_{\mathrm{f}}=3.36 	imes 10^5 \mathrm{~J} / \mathrm{kg}$
The mass of water is $\mathrm{m}_{\mathrm{w}}=100 \mathrm{~kg}$,
The initial temperature of water is $\mathrm{T}_{\mathrm{w}}=25^{\circ} \mathrm{C}$,
The specific heat capacity of water is $\mathrm{c}_{\mathrm{w}}=4200 \frac{\mathrm{~J}}{\mathrm{~kg} \cdot{ }^{\circ} \mathrm{C}}$
Let the final equilibrium temperature is $\mathrm{T}_{\mathrm{f}}$.
The heat gained by ice is for three process,
$Q_{	ext {gained }}=Q_1$ (Heating ice to $\left.0^{\circ} \mathrm{C}ight)+Q_2$ (Melting ice at $\left.0^{\circ} \mathrm{C}ight)+Q_3$ (Heating the melted ice (water) to $T_f$ )
$$\Rightarrow $$ $\mathrm{Q}_1=\mathrm{m}_{\mathrm{i}} \mathrm{c}_{\mathrm{i}} \Delta \mathrm{T}=10 	imes 2100 	imes(0-(-10))=210000 \mathrm{~J}$
$$\Rightarrow $$ $\mathrm{Q}_2=\mathrm{m}_{\mathrm{i}} \mathrm{L}_{\mathrm{f}}=10 	imes 3.36 	imes 10^5=3360000 \mathrm{~J}$
$$\Rightarrow $$ $\mathrm{Q}_3=\mathrm{m}_{\mathrm{i}} \mathrm{c}_{\mathrm{w}}\left(\mathrm{T}_{\mathrm{f}}-0ight)=10 	imes 4200 	imes \mathrm{T}_{\mathrm{f}}=42000 \mathrm{~T}_{\mathrm{f}}$
$$ Q_{	ext {gained }}=210000+3360000+42000 \mathrm{~T}_{\mathrm{f}}=3570000+42000 \mathrm{~T}_{\mathrm{f}} $$
The water cools from $25^{\circ} \mathrm{C}$ to $\mathrm{T}_{\mathrm{f}}$. So the heat lost by water is,
$$ \mathrm{Q}_{	ext {loss }}=\mathrm{m}_{\mathrm{w}} \mathrm{c}_{\mathrm{w}}\left(25-\mathrm{T}_{\mathrm{f}}ight)=100 	imes 4200 	imes\left(25-\mathrm{T}_{\mathrm{f}}ight)=420000\left(25-\mathrm{T}_{\mathrm{f}}ight) $$
$$\Rightarrow $$ $$ \mathrm{Q}_{	ext {loss }}=10500000-420000 \mathrm{~T}_{\mathrm{f}} $$
Using conservation of energy, $\mathrm{Q}_{	ext {gain }}=\mathrm{Q}_{	ext {loss }}$
$$ 3570000+42000 \mathrm{~T}_{\mathrm{f}}=10500000-420000 \mathrm{~T}_{\mathrm{f}} $$
$$\Rightarrow $$ $462000 \mathrm{~T}_{\mathrm{f}}=6930000$
$$\Rightarrow $$ $$ \mathrm{T}_{\mathrm{f}}=\frac{6930000}{462000}=15^{\circ} \mathrm{C} $$
So the decrease in the temperature of the water:
$$ \Delta \mathrm{T}_{	ext {water }}=\mathrm{T}_{	ext {initial }}-\mathrm{T}_{\mathrm{f}}=25^{\circ} \mathrm{C}-15^{\circ} \mathrm{C}=10^{\circ} \mathrm{C} $$
Therefore, the decrement to the temperature of water is $10^{\circ} \mathrm{C}$.
Hence, the correct option is (B).

Answer

15

Answer

10

Answer

6.67

Answer

11.6

Thermodynamics & KTG

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