In an industrial process 10 kg of water per hour is to be heated from 20°C to 80°C. To do this steam at 150°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°C. how many kg of steam is required per hour. (Specific heat of steam = 1 calorie per gm°C, Latent heat of vaporisation = 540 cal/gm

The correct answer is 1 kg.

To find the mass of steam required per hour, we can apply the principle of calorimetry, which states that the heat gained by the water must equal the heat lost by the steam in a closed system (assuming no heat loss to the surroundings or the copper coil).

Step 1: Calculate the heat gained by the water.
The mass of water to be heated per hour is:
$m_w=10\text{kg}=10,000\text{g}$

The temperature of the water is raised from $20°\text{C}$ to $80°\text{C}$. Thus, the change in temperature is:
$\Delta T_w=80°\text{C}-20°\text{C}=60°\text{C}$

The specific heat capacity of water is:
$s_w=1\text{cal/(g}°\text{C)}$

The heat gained by the water ($Q_{\text{gained}}$) per hour is calculated using the formula:
$Q_{\text{gained}}=m_w\times s_w\times \Delta T_w$

Substituting the values:
$Q_{\text{gained}}=10,000\text{g}\times 1\text{cal/(g}°\text{C)}\times 60°\text{C}=600,000\text{calories}$

Step 2: Calculate the heat lost by the steam.
Let $m_s$ be the mass of steam required in grams per hour. The steam undergoes three stages of cooling and phase change:

1. Cooling of steam from $150°\text{C}$ to its condensation point of $100°\text{C}$:
$Q_1=m_s\times s_{\text{steam}}\times (150-100)$
Given the specific heat of steam is $1\text{cal/(g}°\text{C)}$:
$Q_1=m_s\times 1\times 50=50m_s\text{calories}$

2. Condensation of steam at $100°\text{C}$ to water at $100°\text{C}$:
$Q_2=m_s\times L$
Given the latent heat of vaporisation $L=540\text{cal/g}$:
$Q_2=m_s\times 540=540m_s\text{calories}$

3. Cooling of the condensed water from $100°\text{C}$ to $90°\text{C}$:
$Q_3=m_s\times s_w\times (100-90)$
$Q_3=m_s\times 1\times 10=10m_s\text{calories}$

The total heat lost by the steam ($Q_{\text{lost}}$) is:
$Q_{\text{lost}}=Q_1+Q_2+Q_3$
$Q_{\text{lost}}=50m_s+540m_s+10m_s=600m_s\text{calories}$

Step 3: Equate heat gained to heat lost.
Using the heat balance equation:
$Q_{\text{gained}}=Q_{\text{lost}}$
$600,000=600m_s$

Solving for $m_s$:
$m_s=\frac{600,000}{600}=1,000\text{g}$

Converting the mass to kilograms:
$m_s=1\text{kg}$

Therefore, 1 kg of steam is required per hour.