Question

As shown in the figure below, you have a system including a Iron cup with mass Mcup = 0.462 kg. The cup contains water with mass Mwater = 0.230 kg. Both the cup and water are initially at a temperature of 51.8 degrees Celsius. You drop ice into the system. The ice has an initial temperature of -24.5 degrees Celsius and a mass of Mice = 0.5293 kg. The system will reach equilibrium with only some of the ice having melted and settle at the final temperature of T = 0 degrees Celsius. Calorimetry_SomeMelts Find the mass of ice that melted, Mmelt = kg Material c (J/(kg K)) Aluminum 897 Beryllium 1820 Cadmium 231 Iron 412 Lead 129 Polyethylene 2303 Steel 466 Uranium 116 Water c (J/(kg K)) ice 2090 water 4186 steam 2010 Water L (J/kg) Fusion 334000 Vaporization 2230000

Answer 1

To find the mass of ice that melted, we can use the principle of conservation of energy and the specific heat equation:

`\[ Q_{\text{cup}} + Q_{\text{water}} + Q_{\text{ice}} = 0 \]`

where
`\( Q \)` is the heat absorbed or released, and the specific heat equation is
`\( Q = mc\Delta T \)`, where
`\( m \)` is the mass,
`\( c \)` is the specific heat, and
`\( \Delta T \)` is the temperature change.

Let's calculate each term:

1. Heat gained by the cup
`(\( Q_{\text{cup}} \))`:

`\[ Q_{\text{cup}} = m_{\text{cup}}c_{\text{iron}}\Delta T_{\text{cup}} \]`

2. Heat gained by the water
`(\( Q_{\text{water}} \))`:

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

3. Heat lost by the melted ice
`(\( Q_{\text{ice}} \))`:

`\[ Q_{\text{ice}} = m_{\text{melt}}L_{\text{fusion}} \]`

Setting
`\( Q_{\text{cup}} + Q_{\text{water}} + Q_{\text{ice}} = 0 \)`, we can solve for
`\( m_{\text{melt}} \)`.

To calculate the mass of ice that melted
`(\(m_{\text{melt}}\))`, we'll use the principle of conservation of energy and the specific heat equation:

`\[ Q_{\text{cup}} + Q_{\text{water}} + Q_{\text{ice}} = 0 \]`

1. Heat gained by the cup
`(\(Q_{\text{cup}}\))`:

`\[ Q_{\text{cup}} = m_{\text{cup}}c_{\text{iron}}\Delta T_{\text{cup}} \]`

`\[ Q_{\text{cup}} = (0.462 \ \text{kg})(412 \ \text{J/(kg K)})(-51.8 \ \text{°C}) \]`

2. Heat gained by the water
`(\(Q_{\text{water}}\))`:

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

`\[ Q_{\text{water}} = (0.23 \ \text{kg})(4186 \ \text{J/(kg K)})(-51.8 \ \text{°C}) \]`

3. Heat lost by the melted ice
`(\(Q_{\text{ice}}\))`:

`\[ Q_{\text{ice}} = m_{\text{melt}}L_{\text{fusion}} \]`

`\[ Q_{\text{ice}} = m_{\text{melt}}(334000 \ \text{J/kg}) \]`

Now, plug these values into the conservation of energy equation and solve for
`\(m_{\text{melt}}\)`:

`\[ (0.462 \cdot 412 \cdot -51.8) + (0.23 \cdot 4186 \cdot -51.8) + (m_{\text{melt}} \cdot 334000) = 0 \]`

After solving, we get
`\(m_{\text{melt}} \approx 0.084 \ \text{kg}\)`. Therefore, approximately 0.084 kg of ice melted.