To solve the problem, we need to calculate the time required for the water in the pitcher to cool from \(30^\circ C\) to \(28^\circ C\) due to evaporation. We will follow these steps:
### Step 1: Calculate the heat required to cool the water
The heat lost by the water when it cools can be calculated using the formula:
\[
Q = m \cdot s \cdot \Delta T
\]
where:
- \(m\) = mass of water = \(9.5 \, \text{kg} = 9500 \, \text{g}\)
- \(s\) = specific heat of water = \(1 \, \text{kcal/kg}^\circ C = 1000 \, \text{cal/g}^\circ C\)
- \(\Delta T\) = change in temperature = \(30^\circ C - 28^\circ C = 2^\circ C\)
Substituting the values:
\[
Q = 9500 \, \text{g} \cdot 1000 \, \text{cal/g}^\circ C \cdot 2^\circ C
\]
\[
Q = 9500 \cdot 1000 \cdot 2 = 19000000 \, \text{cal}
\]
Thus, the heat required to cool the water is:
\[
Q = 19000 \, \text{cal}
\]
### Step 2: Calculate the heat lost due to evaporation
The heat lost due to evaporation can be calculated using the formula:
\[
Q_{\text{evap}} = m_{\text{evap}} \cdot L_f
\]
where:
- \(m_{\text{evap}}\) = mass of water evaporated (in grams)
- \(L_f\) = latent heat of vaporization = \(580 \, \text{cal/g}\)
Since the pitcher loses \(1 \, \text{g}\) of water per minute, the mass evaporated in \(t\) minutes is:
\[
m_{\text{evap}} = 1 \, \text{g/min} \cdot t \, \text{min} = t \, \text{g}
\]
Thus, the heat lost due to evaporation becomes:
\[
Q_{\text{evap}} = t \cdot 580 \, \text{cal}
\]
### Step 3: Set the heat lost equal to the heat required
Since the heat lost due to evaporation must equal the heat required to cool the water, we set the two equations equal to each other:
\[
Q = Q_{\text{evap}}
\]
\[
19000 \, \text{cal} = t \cdot 580 \, \text{cal}
\]
### Step 4: Solve for \(t\)
Rearranging the equation to solve for \(t\):
\[
t = \frac{19000 \, \text{cal}}{580 \, \text{cal}} \approx 32.76 \, \text{min}
\]
### Step 5: Conclusion
Thus, the time required for the water in the pitcher to cool from \(30^\circ C\) to \(28^\circ C\) is approximately \(32.76\) minutes.
