Question

What mass of steam initially at l30°C is needed to warm 200 g of water in a 100 g glass container from 20.0°C to 50.0°C. Take specific heat capacity of steam to be 2.01 x l0^3 J/ kg.°C and that of water to be 4.19 x 10^3 J/ kg.°C.

Hint: The steam loses energy in three stages. In the first stage, the steam is cooled to 100°C where ms is the mass of the steam. In the second stage, the steam is converted to water without the change in temperature. In the third stage, the water created from steam is reduced to water at 50°C.

Answer 1

Answer on Question #80269, Physics / Other

What mass of steam initially at $130{}^{\circ}\mathrm{C}$ is needed to warm $200\mathrm{g}$ of water in a $100\mathrm{g}$ glass container from $20.0{}^{\circ}\mathrm{C}$ to $50.0{}^{\circ}\mathrm{C}$. Take specific heat capacity of steam to be $2.01 \times 10^{3} \mathrm{~J} / \mathrm{kg} \cdot {}{}^{\circ}\mathrm{C}$ and that of water to be $4.19 \times 10^{3} \mathrm{~J} / \mathrm{kg} \cdot {}{}^{\circ}\mathrm{C}$.

Hint: The steam loses energy in three stages. In the first stage, the steam is cooled to $100{}^{\circ}\mathrm{C}$ where ms is the mass of the steam. In the second stage, the steam is converted to water without the change in temperature. In the third stage, the water created from steam is reduced to water at $50{}^{\circ}\mathrm{C}$.

Solution:

This is a heat transfer problem where we know the final temperature of the glass and water combination. Since there might be a phase transition at the water/steam boundary, we have to do this problem in 3 parts: part 1 is cooling the steam, part 2 is condensing, and part 3 is further cooling of the water:

1) Steam cooling from $130{}^{\circ}\mathrm{C}$ to $100{}^{\circ}\mathrm{C}$:

$$Q _ {1} = m _ {\text {steam}} c _ {\text {steam}} \cdot \left(T _ {f} - T _ {i}\right) = 3 0 \cdot m _ {\text {steam}} c _ {\text {steam}}$$

2) Steam condensing:

$$Q _ {2} = m _ {\text {steam}} L _ {v}$$

3) Water (was steam) cooling from $100{}^{\circ}\mathrm{C}$ to $50{}^{\circ}\mathrm{C}$:

$$Q _ {3} = m _ {\text {steam}} c _ {w} \cdot \left(T _ {f} - T _ {i}\right) = 5 0 \cdot m _ {\text {steam}} c _ {w}$$

The total heat liberated by the steam is then the sum of the above:

$$Q _ {o u t} = Q _ {1} + Q _ {2} + Q _ {3}$$

The heat gained by the water/glass combination going from $20{}^{\circ}\mathrm{C}$ to $50{}^{\circ}\mathrm{C}$ is:

$$\begin{array}{l} Q _ {i n} = m _ {w} c _ {w} \cdot \left(T _ {f} - T _ {i}\right) + m _ {\text {glass}} c _ {\text {glass}} \cdot \left(T _ {f} - T _ {i}\right) \\ = \left(m _ {w} c _ {w} + m _ {\text {glass}} c _ {\text {glass}}\right) \cdot 3 0 {}^ {\circ} \mathrm {C} \\ \end{array}$$

These two heats must be equal:

$$\begin{array}{l} Q _ {o u t} = Q _ {i n} \\ m _ {\text {steam}} \left(3 0 \cdot c _ {\text {steam}} + L _ {v} + 5 0 \cdot c _ {w}\right) = \left(m _ {w} c _ {w} + m _ {\text {glass}} c _ {\text {glass}}\right) \cdot 3 0 \\ \end{array}$$ $$m_{steam} = \frac{(m_w c_w + m_{glass} c_{glass}) \cdot 30}{30 \cdot c_{steam} + L_v + 50 \cdot c_w}$$

The specific heats of glass is 837 J/kg.

So,

$$m_{steam} = \frac{(0.2 \times 4190 + 0.1 \times 837) \cdot 30}{30 \cdot 2010 + 2.26 \times 10^6 + 50 \cdot 4190} = 0.0109\ \mathrm{kg} \approx 10.9\ \mathrm{g}$$

Answer: 10.9 g

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