Question

In an industrial process 10 kg of water per hour is to be heated from ${20}^{\circ }Cto{80}^{\circ }C.$ 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 ${90}^{\circ }C.$ how many kg of steam is required per hour.
(Specific heat of steam = 1 calorie per gm°C, Latent heat of vaporization = 540 cal/gm)

• A. 1 gm
• B. 1 kg
• C. 10 gm
• D. 10 kg

Answer 1

The correct option is B 1 kg

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