Question

(b) What mass of steam at 130 °C must be condensed on a 0.100 kg glass cup to warm the cup and the 0.200 kg of water it contains from 20.0 °C to 50.0 °C? [5] [Specific heat capacity of steam, cs= 2.01 x 10³ J.kg¹.K¹] [Specific heat capacity of water, cw = 4.19 x 10³ J.kg¹.K¹] 7 [Specific heat capacity of glass, cg = 8.37 x 102 J.kg¹.K¹] [Latent of vaporisation, L = 2.26 x 10° J.kg¹¹]​

Answer 1

No mass of steam is needed to warm the cup and water from 20.0 °C to 50.0 °C because all the required heat comes from the condensation of the steam.

To solve this problem, we need to consider the heat transferred to the glass cup and the water in it. The heat transferred is given by the equation:

`\[ Q = mc\Delta T \]`

where Q is the heat transferred, m is the mass, c is the specific heat capacity, and
`\( \Delta T \)` is the temperature change.

The total heat transferred will be the sum of the heat needed to raise the temperature of the glass cup, the water, and the heat needed to condense the steam.

`\[ Q_{\text{total}} = Q_{\text{glass}} + Q_{\text{water}} + Q_{\text{steam}} \]`

1. Heat for the glass cup:

`\[ Q_{\text{glass}} = m_{\text{glass}} \cdot c_{\text{glass}} \cdot \Delta T_{\text{glass}} \]`

2. Heat for the water:

`\[ Q_{\text{water}} = m_{\text{water}} \cdot c_{\text{water}} \cdot \Delta T_{\text{water}} \]`

3. Heat for the steam to condense:

`\[ Q_{\text{steam}} = m_{\text{steam}} \cdot L \]`

Now, let's calculate each part step by step.

Given values:

`\[ c_{\text{steam}} = 2.01 * 10^3 \, \text{J/kg} \cdot \text{K} \]`

`\[ c_{\text{water}} = 4.19 * 10^3 \, \text{J/kg} \cdot \text{K} \]`

`\[ c_{\text{glass}} = 8.37 * 10^2 \, \text{J/kg} \cdot \text{K} \]`

`\[ L = 2.26 * 10^6 \, \text{J/kg} \]`

Given masses and temperature changes:

`\[ m_{\text{glass}} = 0.1 \, \text{kg} \]`

`\[ m_{\text{water}} = 0.2 \, \text{kg} \]`

`\[ \Delta T_{\text{glass}} = \Delta T_{\text{water}} = 50 \, \text{\textdegree C} - 20 \, \text{\textdegree C} = 30 \, \text{\textdegree C} \]`

1. Heat for the glass cup:

`\[ Q_{\text{glass}} = 0.1 \, \text{kg} \cdot 8.37 * 10^2 \, \text{J/kg} \cdot \text{K} \cdot 30 \, \text{K} \]`

`\[ Q_{\text{glass}} = 2.511 * 10^4 \, \text{J} \]`

2. Heat for the water:

`\[ Q_{\text{water}} = 0.2 \, \text{kg} \cdot 4.19 * 10^3 \, \text{J/kg} \cdot \text{K} \cdot 30 \, \text{K} \]`

`\[ Q_{\text{water}} = 2.514 * 10^4 \, \text{J} \]`

3. Heat for the steam to condense:

`\[ Q_{\text{steam}} = m_{\text{steam}} \cdot L = ? \]`

Now, let's calculate
`\( m_{\text{steam}} \)` using the fact that the total heat transferred is equal to the sum of the individual heats:

`\[ Q_{\text{total}} = Q_{\text{glass}} + Q_{\text{water}} + Q_{\text{steam}} \]`

`\[ Q_{\text{total}} = 2.511 * 10^4 \, \text{J} + 2.514 * 10^4 \, \text{J} + Q_{\text{steam}} \]`

`\[ Q_{\text{total}} = 5.025 * 10^4 \, \text{J} + Q_{\text{steam}} \]`

Now, subtract the heat for the glass and water from the total heat to find the heat needed to condense the steam:

`\[ Q_{\text{steam}} = Q_{\text{total}} - (Q_{\text{glass}} + Q_{\text{water}}) \]`

`\[ Q_{\text{steam}} = 5.025 * 10^4 \, \text{J} - (2.511 * 10^4 \, \text{J} + 2.514 * 10^4 \, \text{J}) \]`

`\[ Q_{\text{steam}} = 5.025 * 10^4 \, \text{J} - 5.025 * 10^4 \, \text{J} \]`

`\[ Q_{\text{steam}} = 0 \]`

Since the heat needed to condense the steam is zero, it means that all the heat required to warm the glass cup and water comes from the condensation of the steam.

Therefore, the mass of steam needed is
`\( m_{\text{steam}} = 0 \)`.

In conclusion, no mass of steam is needed to warm the cup and water from 20.0 °C to 50.0 °C because all the required heat comes from the condensation of the steam.