Question

Steam at 100°C is passed through 1 kg of water at 0°C Contained in a Calorimeter of mass 0.2 kg. Calculate the mass of steam Required to Revise Temperature to 20°c

• A. 32.86 g
• B. 45.21 g
• C. 21.55 g
• D. 50.00 g

Answer 1

Answer: (A)

32.86 g

Answer 2

The problem involves a calorimetry scenario where steam at $100^\circ\text{C}$ is introduced into water and a calorimeter, both initially at $0^\circ\text{C}$. The goal is to find the mass of steam needed to raise the final temperature to $20^\circ\text{C}$.

Assumptions:

• The initial temperature of water and calorimeter is $0^\circ\text{C}$.
• The specific heat capacity of water ($c_w$) is $4186 \, \text{J/kg}^\circ\text{C}$.
• The specific heat capacity of the calorimeter material is assumed to be that of copper, $c_c = 385 \, \text{J/kg}^\circ\text{C}$.
• The latent heat of vaporization of steam ($L_v$) is $2.26 \times 10^6 \, \text{J/kg}$.
• The steam, upon condensation, becomes water at $100^\circ\text{C}$, and then cools down to the final temperature.

Given:

• Mass of water ($m_w$): $1 \, \text{kg}$
• Initial temperature ($T_{initial}$): $0 \, ^\circ\text{C}$
• Mass of calorimeter ($m_c$): $0.2 \, \text{kg}$
• Temperature of steam ($T_{steam}$): $100 \, ^\circ\text{C}$
• Final temperature ($T_{final}$): $20 \, ^\circ\text{C}$

Calculations:

1. Heat gained by water: $Q_{gain, w} = m_w \times c_w \times (T_{final} - T_{initial})$ $Q_{gain, w} = 1 \, \text{kg} \times 4186 \, \text{J/kg}^\circ\text{C} \times (20 \, ^\circ\text{C} - 0 \, ^\circ\text{C})$ $Q_{gain, w} = 1 \times 4186 \times 20 = 83720 \, \text{J}$

2. Heat gained by calorimeter: $Q_{gain, c} = m_c \times c_c \times (T_{final} - T_{initial})$ $Q_{gain, c} = 0.2 \, \text{kg} \times 385 \, \text{J/kg}^\circ\text{C} \times (20 \, ^\circ\text{C} - 0 \, ^\circ\text{C})$ $Q_{gain, c} = 0.2 \times 385 \times 20 = 1540 \, \text{J}$

3. Total heat gained: $Q_{gain, total} = Q_{gain, w} + Q_{gain, c} = 83720 \, \text{J} + 1540 \, \text{J} = 85260 \, \text{J}$

4. Heat lost by steam: Let $m_s$ be the mass of steam. The steam first condenses at $100^\circ\text{C}$ and then cools down to $20^\circ\text{C}$.

   • Heat lost during condensation: $Q_{condense} = m_s \times L_v = m_s \times (2.26 \times 10^6 \, \text{J/kg})$
   • Heat lost during cooling of condensed steam (water) from $100^\circ\text{C}$ to $20^\circ\text{C}$: $Q_{cool} = m_s \times c_w \times (100^\circ\text{C} - 20^\circ\text{C}) = m_s \times 4186 \, \text{J/kg}^\circ\text{C} \times 80 \, ^\circ\text{C}$ $Q_{cool} = m_s \times 334880 \, \text{J/kg}$

   Total heat lost by steam: $Q_{lost, s} = Q_{condense} + Q_{cool} = m_s \times (2.26 \times 10^6) + m_s \times 334880$ $Q_{lost, s} = m_s \times (2260000 + 334880) \, \text{J/kg}$ $Q_{lost, s} = m_s \times 2594880 \, \text{J/kg}$

5. Equating heat lost and gained: $Q_{lost, s} = Q_{gain, total}$ $m_s \times 2594880 \, \text{J/kg} = 85260 \, \text{J}$ $m_s = \frac{85260}{2594880} \, \text{kg}$ $m_s \approx 0.032856 \, \text{kg}$

   Converting to grams: $m_s \approx 0.032856 \times 1000 \, \text{g} \approx 32.86 \, \text{g}$

The mass of steam required is approximately $32.86 \, \text{g}$.