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

Therefore, the mass of steam that must be condensed is approximately
`\(0.037 \, \text{kg}\).`

To solve this problem, we need to consider the heat transfer involved in two processes: first, raising the temperature of the glass and water, and second, condensing steam into water.

1. Heat transfer to raise the temperature of the glass and water:

`\[ Q_1 = m_{\text{glass}} \cdot c_{\text{glass}} \cdot \Delta T_{\text{glass}} + m_{\text{water}} \cdot c_{\text{water}} \cdot \Delta T_{\text{water}} \]`

Where:

`\( m_{\text{glass}} \) is the mass of the glass,`

`\( c_{\text{glass}} \) is the specific heat capacity of the glass,`

`\( \Delta T_{\text{glass}} \)` is the change in temperature of the glass,

`\( m_{\text{water}} \)` is the mass of water,

`\( c_{\text{water}} \)` is the specific heat capacity of water,

`\( \Delta T_{\text{water}} \)`is the change in temperature of the water.

2. Heat transfer to condense steam into water:

`\[ Q_2 = m_{\text{steam}} \cdot L \]`

Where:

-
`\( m_{\text{steam}} \)` is the mass of steam,

-
`\( L \)` is the latent heat of vaporization.

The total heat transfer
`(\( Q_{\text{total}} \))` is the sum of
`\( Q_1 \)` and
`\( Q_2 \)`:

`\[ Q_{\text{total}} = Q_1 + Q_2 \]`

Now, let's substitute the given values:

-
`\( m_{\text{glass}} = 0.100 \, \text{kg} \)`

-
`\( c_{\text{glass}} = 8.37 * 10^2 \, \text{J} \cdot \text{kg}^(-1) \cdot \text{K}^(-1) \)`

-
`\( \Delta T_{\text{glass}} = 50.0 \, \text{K} \) (final temperature - initial temperature)`

-
`\( m_{\text{water}} = 0.200 \, \text{kg} \)`

-
`\( c_{\text{water}} = 4.19 * 10^3 \, \text{J} \cdot \text{kg}^(-1) \cdot \text{K}^(-1) \)`

-
`\( \Delta T_{\text{water}} = 50.0 \, \text{K} \) (final temperature - initial temperature)`

-
`\( m_{\text{steam}} \)` is what we're trying to find.

-
`\( L = 2.26 * 10^6 \, \text{J} \cdot \text{kg}^(-1) \)Now, let's calculate \( Q_1 \)`:

`\[ Q_1 = (0.100 \, \text{kg} \cdot 8.37 * 10^2 \, \text{J} \cdot \text{kg}^(-1) \cdot \text{K}^(-1) \cdot 50.0 \, \text{K}) + (0.200 \, \text{kg} \cdot 4.19 * 10^3 \, \text{J} \cdot \text{kg}^(-1) \cdot \text{K}^(-1) \cdot 50.0 \, \text{K}) \]`

`\[ Q_1 = 4.185 * 10^4 \, \text{J} + 4.185 * 10^4 \, \text{J} \]`

`\[ Q_1 = 8.37 * 10^4 \, \text{J} \]`

Now, let's find
`\( Q_2 \)`:

`\[ Q_2 = m_{\text{steam}} \cdot L \]`

`\[ 8.37 * 10^4 \, \text{J} + m_{\text{steam}} \cdot 2.26 * 10^6 \, \text{J} \cdot \text{kg}^(-1) \]`

Solving for
`\( m_{\text{steam}} \)`:

`\[ m_{\text{steam}} = \frac{8.37 * 10^4 \, \text{J}}{2.26 * 10^6 \, \text{J} \cdot \text{kg}^(-1)} \]`

`\[ m_{\text{steam}} \approx 0.037 \, \text{kg} \]`

Therefore, the mass of steam that must be condensed is approximately
`\(0.037 \, \text{kg}\).`