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A backpacker wants to carry enough fuel to heat [tex]2.6 \, \text{kg}[/tex] of water from [tex]29^{\circ} \text{C}[/tex] to [tex]100.0^{\circ} \text{C}[/tex].

Part A

If the fuel she carries produces [tex]36 \, \text{kJ}[/tex] of heat per gram when it burns, how much fuel should she carry? The specific heat capacity of water is [tex]4.184 \, \text{J} /\left( \text{g} \cdot {}^{\circ} \text{C} \right)[/tex]. (For the sake of simplicity, assume that the transfer of heat is [tex]100\%[/tex] efficient.)

Express your answer in grams to two significant figures.


Sagot :

To determine the amount of fuel that the backpacker needs to carry in order to heat 2.6 kg of water from 29°C to 100°C, we will follow these steps:

1. Convert the mass of water from kilograms to grams:
[tex]\[ \text{Mass of water} = 2.6 \, \text{kg} \times 1000 \, \text{g/kg} = 2600 \, \text{g} \][/tex]

2. Determine the temperature change (\(\Delta T\)):
[tex]\[ \Delta T = 100^\circ C - 29^\circ C = 71^\circ C \][/tex]

3. Calculate the amount of heat required (Q) using the formula:
[tex]\[ Q = \text{mass} \times \text{specific heat capacity} \times \Delta T \][/tex]
Given:
[tex]\[ \text{mass} = 2600 \, \text{g}, \quad \text{specific heat capacity} = 4.184 \, \text{J/(g·}^\circ C\text{)} \][/tex]
[tex]\[ Q = 2600 \, \text{g} \times 4.184 \, \text{J/(g·}^\circ C\text{)} \times 71^\circ C = 772366.4 \, \text{J} \][/tex]

4. Convert the heat required to kilojoules:
[tex]\[ Q = 772366.4 \, \text{J} \times \frac{1 \, \text{kJ}}{1000 \, \text{J}} = 772.3664 \, \text{kJ} \][/tex]

5. Determine the amount of fuel needed using the heat per gram of fuel:
[tex]\[ \text{Heat per gram of fuel} = 36 \, \text{kJ/g} \][/tex]
[tex]\[ \text{Fuel needed} = \frac{Q}{\text{Heat per gram of fuel}} = \frac{772.3664 \, \text{kJ}}{36 \, \text{kJ/g}} = 21.454622 \, \text{g} \][/tex]

6. Round the answer to two significant figures:
[tex]\[ \text{Fuel needed} \approx 21 \, \text{g} \][/tex]

So, the backpacker should carry 21 grams of fuel.