To solve the problem step by step, we will use the principle of conservation of energy, which states that the heat lost by the container will be equal to the heat gained by the ice cube.
### Step 1: Define the heat lost by the container
The heat lost by the container can be expressed using the formula:
\[
dq = m \cdot c \cdot dT
\]
where:
- \( m \) is the mass of the container,
- \( c \) is the specific heat of the container, which varies with temperature as \( c = A + BT \),
- \( dT \) is the change in temperature.
### Step 2: Set up the specific heat equation
Given:
- \( A = 100 \, \text{cal/kg.K} \)
- \( B = 2 \times 10^{-2} \, \text{cal/kg.K}^2 \)
- The initial temperature of the container \( T_i = 227^\circ C = 500 \, K \)
- The final temperature \( T_f = 27^\circ C = 300 \, K \)
Thus, the specific heat at temperature \( T \) is:
\[
c = 100 + 0.02T
\]
### Step 3: Calculate the heat lost by the container
The heat lost by the container from \( T_i \) to \( T_f \) can be calculated by integrating:
\[
Q_{\text{lost}} = \int_{T_i}^{T_f} m \cdot (100 + 0.02T) \, dT
\]
This can be simplified as:
\[
Q_{\text{lost}} = m \left[ 100T + 0.01T^2 \right]_{500}^{300}
\]
Calculating this:
\[
Q_{\text{lost}} = m \left[ (100 \cdot 300 + 0.01 \cdot 300^2) - (100 \cdot 500 + 0.01 \cdot 500^2) \right]
\]
\[
= m \left[ (30000 + 900) - (50000 + 2500) \right]
\]
\[
= m \left[ 30900 - 52500 \right]
\]
\[
= m \left[ -21600 \right] \, \text{calories}
\]
### Step 4: Calculate the heat gained by the ice
The ice cube first melts and then warms up to the final temperature. The heat gained by the ice can be calculated in two parts:
1. Heat required to melt the ice:
\[
Q_1 = m_{\text{ice}} \cdot L_f = 0.1 \, \text{kg} \cdot 80000 \, \text{cal/kg} = 8000 \, \text{cal}
\]
2. Heat required to raise the temperature of the melted ice (water) from \( 0^\circ C \) to \( 27^\circ C \):
\[
Q_2 = m_{\text{water}} \cdot c_{\text{water}} \cdot \Delta T = 0.1 \, \text{kg} \cdot 1000 \, \text{cal/kg.K} \cdot 27 \, K = 2700 \, \text{cal}
\]
### Step 5: Total heat gained by the ice
\[
Q_{\text{gained}} = Q_1 + Q_2 = 8000 + 2700 = 10700 \, \text{cal}
\]
### Step 6: Set up the equation for heat lost and gained
According to the conservation of energy:
\[
|Q_{\text{lost}}| = Q_{\text{gained}}
\]
\[
21600m = 10700
\]
### Step 7: Solve for the mass of the container
\[
m = \frac{10700}{21600} \approx 0.495 \, \text{kg}
\]
### Final Answer
The mass of the container is approximately \( 0.495 \, \text{kg} \).
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