How much steam at `100^(@)C` is needed to change 40 g of ice at `-10^(@)C` to water at `20^(@)C` if the ice in a 50 g copper can? Assume that the can maintains the same temperature as the ice and water. Given that specific heat of copper is `S_(Cu)=0.09 "cal"//gm^(@)` and specific heat of ice is `S_("ice")=0.5` cal /gm`.^(@)C` .

Answer

To solve the problem of how much steam at \(100^\circ C\) is needed to change \(40 \, g\) of ice at \(-10^\circ C\) to water at \(20^\circ C\) in a \(50 \, g\) copper can, we will follow these steps:

### Step 1: Calculate the heat required to raise the temperature of ice from \(-10^\circ C\) to \(0^\circ C\).

The formula to calculate heat is:
\[ Q = m \cdot S \cdot \Delta T \]
where:
- \(m\) = mass of ice = \(40 \, g\)
- \(S\) = specific heat of ice = \(0.5 \, \text{cal/g}^\circ C\)
- \(\Delta T\) = change in temperature = \(0 - (-10) = 10^\circ C\)

Calculating the heat:
\[ Q_1 = 40 \, g \cdot 0.5 \, \text{cal/g}^\circ C \cdot 10^\circ C = 200 \, \text{cal} \]

### Step 2: Calculate the heat required to melt the ice at \(0^\circ C\).

The latent heat of fusion of ice is \(80 \, \text{cal/g}\). The heat required to melt the ice is:
\[ Q_2 = m \cdot L_f \]
where:
- \(L_f\) = latent heat of fusion = \(80 \, \text{cal/g}\)

Calculating the heat:
\[ Q_2 = 40 \, g \cdot 80 \, \text{cal/g} = 3200 \, \text{cal} \]

### Step 3: Calculate the heat required to raise the temperature of water from \(0^\circ C\) to \(20^\circ C\).

Now, we need to raise the temperature of the resulting water:
\[ Q_3 = m \cdot S_w \cdot \Delta T \]
where:
- \(S_w\) = specific heat of water = \(1 \, \text{cal/g}^\circ C\)
- \(\Delta T\) = change in temperature = \(20 - 0 = 20^\circ C\)

Calculating the heat:
\[ Q_3 = 40 \, g \cdot 1 \, \text{cal/g}^\circ C \cdot 20^\circ C = 800 \, \text{cal} \]

### Step 4: Calculate the heat required to raise the temperature of the copper can from \(-10^\circ C\) to \(20^\circ C\).

Using the same formula:
\[ Q_4 = m \cdot S_{Cu} \cdot \Delta T \]
where:
- \(m\) = mass of copper = \(50 \, g\)
- \(S_{Cu}\) = specific heat of copper = \(0.09 \, \text{cal/g}^\circ C\)
- \(\Delta T\) = change in temperature = \(20 - (-10) = 30^\circ C\)

Calculating the heat:
\[ Q_4 = 50 \, g \cdot 0.09 \, \text{cal/g}^\circ C \cdot 30^\circ C = 135 \, \text{cal} \]

### Step 5: Calculate the total heat required.

Now, we sum all the heats calculated:
\[ Q_{\text{total}} = Q_1 + Q_2 + Q_3 + Q_4 = 200 + 3200 + 800 + 135 = 4335 \, \text{cal} \]

### Step 6: Calculate the mass of steam required.

The heat released by steam when it condenses is given by:
\[ Q = m_s \cdot L_v \]
where:
- \(L_v\) = latent heat of vaporization of steam = \(540 \, \text{cal/g}\)

Setting the total heat required equal to the heat released by the steam:
\[ 4335 \, \text{cal} = m_s \cdot 540 \, \text{cal/g} \]

Solving for \(m_s\):
\[ m_s = \frac{4335 \, \text{cal}}{540 \, \text{cal/g}} \approx 8.02 \, g \]

### Final Answer:
The mass of steam required is approximately \(8.02 \, g\).

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