Question 505: Thermodynamics & KTG

Question

The heat extracted out of $x$ gram of water initially at $50^{\circ} \mathrm{C}$ to $\operatorname{cool}$ it down to $0^{\circ} \mathrm{C}$ is sufficient to evaporate $(1000-x)$ gram of water also initially at $50^{\circ} \mathrm{C}$. The value of $x$ (closest integer) is $\_\_\_\_$ .
(Take latent heat of water $2256 \mathrm{~kJ} / \mathrm{kg} . \mathrm{K}$, specific heat capacity of water $4200 \mathrm{~J} / \mathrm{kg} . \mathrm{K}$ )

Accepted Answer

When x grams of water cools from $50^{\circ} \mathrm{C}$ to $0^{\circ} \mathrm{C}$, the heat released is:
$$ \mathrm{Q}_1=\mathrm{m}_1 \cdot \mathrm{c} \cdot \Delta \mathrm{~T} $$
Where, $\mathrm{m}_1=\mathrm{xg}=\frac{\mathrm{x}}{1000} \mathrm{~kg}$ is the mass of water, $\mathrm{c}=4200 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}$ is the specific heat capacity of water, and $\Delta \mathrm{T}=50-0=50^{\circ} \mathrm{C}$
Putting the values,
$$ Q_1=\left(\frac{x}{1000}ight) 	imes 4200 	imes 50 $$
$$\Rightarrow $$ $$ Q_1=x 	imes 4.2 	imes 50=210 x J $$
The water to be evaporated is initially at $50^{\circ} \mathrm{C}$. To evaporate it, it must first be heated to $100^{\circ} \mathrm{C}$ and then converted to steam.
Let $\mathrm{m}_2=(1000-\mathrm{x}) \mathrm{g}=\frac{1000-\mathrm{x}}{1000} \mathrm{~kg}$.
Heating from $50^{\circ} \mathrm{C}$ to $100^{\circ} \mathrm{C}$
$$ \mathrm{Q}_{2 \mathrm{a}}=\mathrm{m}_2 \cdot \mathrm{c} \cdot(100-50) $$
$$ Q_{2 a}=\left(\frac{1000-x}{1000}ight) 	imes 4200 	imes 50=(1000-x) 	imes 210 \mathrm{~J} $$
Latent heat of vaporization at $100^{\circ} \mathrm{C}$
$$ \mathrm{Q}_{2 \mathrm{~b}}=\mathrm{m}_2 \cdot \mathrm{~L} $$
Where the latent heat of water is $\mathrm{L}=2256 \mathrm{~kJ} / \mathrm{kg}=2,256,000 \mathrm{~J} / \mathrm{kg}$.
$$ Q_{2 b}=\left(\frac{1000-x}{1000}ight) 	imes 2,256,000=(1000-x) 	imes 2256 J $$
So, the total heat for evaporation is
$$ Q_2=Q_{2 a}+Q_{2 b}=(1000-x) 	imes[210+2256]=(1000-x) 	imes 2466 J $$
Using conservation of heat energy, the total amount of heat gained is equal to total amount of heat rejected.
$$ \mathrm{Q}_1=\mathrm{Q}_2 $$
$$ \begin{gathered} 210 \mathrm{x}=(1000-\mathrm{x}) 2466 \ 210 \mathrm{x}=2,466,000-2466 \mathrm{x} \ 210 \mathrm{x}+2466 \mathrm{x}=2,466,000 \ 2676 \mathrm{x}=2,466,000 \ \mathrm{x}=\frac{2,466,000}{2676} \approx 921.52 \end{gathered} $$
Therefore, the value of x (closest integer) is 922 .

Thermodynamics & KTG

Integer / Numeric Answer Type