Question

In an industrial process $10kg$ of water per hour is to be heated from ${20}^{\circ }C\text{to}{80}^{\circ }C$. To do this steam at ${150}^{\circ }C$ is passed from a boiler into a copper coil immersed in water. The steam condenses in the coil and is returned to the boiler as water at ${90}^{\circ }C$ then ___ $kg$ of steam is required per hour.

(Given: Specific heat of steam $=1$ calorie per g ${}^{\circ }C$, latent heat of vaporization $=540cal{g}^{-1}$)

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is A $1$

Heat required by $10kg$ water to change its temperature from ${20}^{\circ }C\text{to}{80}^{\circ }C$ in one hour is,
$Q_1=(mc\Delta T{)}_{water}$
where, $m$ is the mass of water, $c$ is the specific heat of water and $\Delta T$ is the change in temperature of the water.
Given, $c=1cal$, $m=10kg=10\times {10}^{3}g$
$Q_1=(10\times {10}^3)\times 1\times (80-20)$
$=600\times {10}^3\text{calorie}$
In condensation,
(i) Steam release heat when it looses it's temperature from ${150}^{\circ }C\text{to}{100}^{\circ }C$, i.e., $\left[m{c}_{steam\Delta T}\right]$, where, where, $m$ is the mass of steam, $c_{steam}$ is the specific heat of steam and $\Delta T$ is the change in temperature of the steam.
(ii) At ${100}^{\circ }C$, it converts into water and gives the latent heat, i.e., $\left[mL\right]$, where, $m$ is the mass of ice and $L$ is the latent heat of vaporization.
(iii) Water release heat when it looses it's temperature from ${100}^{\circ }C\text{to}{90}^{\circ }C$, i.e., $\left[m{c}_{water}\Delta T\right]$, where, $m$ is the mass of water, $c_{water}$ is the specific heat of water and $\Delta T$ is the change in temperature.
If $m$ gram of steam condenses per hour, then heat released by steam in converting water to ${90}^{\circ }C$,
$Q_2=(mc\Delta T{)}_{steam}+m{L}_{steam}$$+(ms\Delta T{)}_{water}$
$=m[1\times (150-100)+540+$$1\times (100-90)]$
$=600m\text{calorie}$
According to problem $Q_1=Q_2\Rightarrow 600m=600\times {10}^3\Rightarrow m={10}^{3}g=1kg$