Question

In an industrial process 10kg of water per hour is to be heated from 20^oC to 80^oC. To do this, steam at 150^oC 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^oC, how many kg of steam is required per hour?

(Take, specific heat of steam=1 cal per g^oC and latent heat of vaporisation = 540 cal g^-1)

• A. 1g
• B. 1kg
• C. 10g
• D. 10kg

Answer 1

The correct option is B

1kg

Step :1 Given data

Mass of water, $m_w=10kg={10}^{4}g$

The initial temperature of the water, $T_1={20}^{o}C$

The final temperature of the water, $T_2={80}^{o}C$

The initial temperature of steam, $T_1={150}^{o}C$

The final temperature of steam, $T_2={90}^{o}C$

Specific heat of steam, $c_s=1\frac{cal}{g{}^{o}C}$

Latent heat of vaporization, $L=540\frac{cal}{g}$

Step:2 Finding the amount of heat absorbed

Heat energy $Q=mc∆T$

Heat absorbed by water to raise the temperature from 20 ^oC to 80 ^oC is $Q_a=m_w{c}_w∆T$

$\Rightarrow Q_a={10}^{4}g\times 1\frac{cal}{g{}^{o}C}\times \left({80}^{o}C-{20}^{o}C\right)$

$={10}^4\times 60$ $=6\times {10}^{5}cal$

Step :3 Finding the amount of heat released

The heat released by steam to convert into the water from 150 ^oC to 90 ^oC is

$Q_r=m_s{c}_s∆T+m_{s}L+m_s{c}_w∆T$

$Q_r=m_{s}g\times 1\frac{cal}{g{}^{o}C}\times \left({150}^{o}C-100{}^{o}C\right)+m_{s}g\times 540\frac{cal}{g}+m_{s}g\times 1\frac{cal}{g{}^{o}C}\times \left({100}^{o}C-{90}^{o}C\right)$

$=m_s\times 50+m_s\times 540+m_s\times 10$

$=50m_s+540m_s+10m_s=600m_{s}cal$

Step:4 Finding the mass of ice required

According to the principle of calorimetry, the heat released by a body = heat absorbed by a body

Therefore, $Q_r=Q_a$ $\Rightarrow 600m_{s}cal=6\times {10}^{5}cal$

$\Rightarrow m_s=\frac{6\times {10}^5}{600}=1000g=1kg$

So, the mass of steam required is 1 kg. Thus B is the correct option.