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Let's go through the problem step by step.

### Given Data:
1. Initial temperature of water, \( T_{\text{initial}} = 10^\circ \text{C} \).
2. Volume of water, \( V = 11 \, \text{L} = 0.011 \, \text{m}^3 \) (since \( 1 \, \text{L} = 0.001 \, \text{m}^3 \)).
3. Mass of fuel burned, \( m_{\text{fuel}} = 0.08 \, \text{kg} \).
4. Efficiency of the heating system, \( \eta = 49\% = 0.49 \).
5. Heat of combustion of fuel, \( q_{\text{fuel}} = 29 \, \text{MJ/kg} = 29,000,000 \, \text{J/kg} \).
6. Specific heat capacity of water, \( c_{\text{water}} = 4200 \, \text{J/(kg} \cdot \text{°C)} \).
7. Density of water, \( \rho = 1000 \, \text{kg/m}^3 \).

### Step 1: Calculate the Mass of Water
\[
m_{\text{water}} = \rho \cdot V = 1000 \, \text{kg/m}^3 \times 0.011 \, \text{m}^3 = 11 \, \text{kg}
\]

### Step 2: Calculate the Total Energy Released by the Fuel
\[
Q_{\text{total}} = m_{\text{fuel}} \times q_{\text{fuel}} = 0.08 \, \text{kg} \times 29,000,000 \, \text{J/kg} = 2,320,000 \, \text{J}
\]

### Step 3: Calculate the Useful Energy (Account for Efficiency)
Since the efficiency is 49%, only 49% of \( Q_{\text{total}} \) is used to heat the water:
\[
Q_{\text{useful}} = \eta \times Q_{\text{total}} = 0.49 \times 2,320,000 \, \text{J} = 1,136,800 \, \text{J}
\]

### Step 4: Set Up the Heat Transfer Equation
The energy required to raise the temperature of the water is given by:
\[
Q_{\text{useful}} = m_{\text{water}} \times c_{\text{water}} \times (T_{\text{final}} - T_{\text{initial}})
\]
Substitute the values:
\[
1,136,800 = 11 \times 4200 \times (T_{\text{final}} - 10)
\]

### Step 5: Solve for \( T_{\text{final}} \)
1. Simplify the equation:
   \[
   1,136,800 = 46200 \times (T_{\text{final}} - 10)
   \]
2. Divide both sides by 46200:
   \[
   T_{\text{final}} - 10 = \frac{1,136,800}{46200} \approx 24.6
   \]
3. Add 10 to both sides:
   \[
   T_{\text{final}} \approx 24.6 + 10 = 34.6
   \]

### Answer (Rounded to Whole Number):
\[
T_{\text{final}} \approx 35^\circ \text{C}
\]

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Would you like further explanation on any step?

### Related Questions
1. How would the final temperature change if the efficiency was increased to 60%?
2. What would be the result if the volume of water was doubled?
3. How much fuel would be required to reach a temperature of \(50^\circ \text{C}\)?
4. What would happen if the specific heat capacity of the substance being heated was lower than that of water?
5. How would a decrease in the initial temperature affect the final temperature?

### Tip
When dealing with efficiency, always remember to calculate the useful energy by multiplying the total energy by the efficiency rate before applying it to your heat transfer equations.

Определи, до какой температуры нагрелась находящаяся в сосуде вода объёмом 11 л и температурой 10°C, если для её нагревания было израсходовано 0,08 кг горючего. КПД нагревательной установки — 49%. Справочные данные: удельная теплота сгорания горючего — 29 МДж/кг, удельная теплоёмкость воды — 4200 Дж/(кг·°C), плотность воды — 1000 кг/м³.

Learn how to determine the final temperature of water heated in a container with an initial temperature of 10°C and a volume of 11 L. This thermodynamics problem involves calculating the efficiency of the heating system, which is given as 49%, and uses the heat of combustion for fuel and specific heat of water. Suitable for high school students.

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