Answer:
66°C
Explanation:
Energy obtained from burning 1.21 pounds of butane gives a fuel value of 10.85 kcal/g
mass of water = 123.0kg
Initial temperature = 18.0°C
Final temperature = α
Energy released from burning butane is used to heat water to the final temperature
Energy released = Q released
Q released = 1.21 * 453.59 * 10.85
= 5954.96J
Energy needed = Q needed
Q needed = 123 * 1 * (α - 18)
= 123(α - 18)
Q needed = Q released
123(α - 18) = 5954.96
α - 18 = 5954.89/123
α - 18 = 48.41
α = 48.41 + 18
α = 66.41
α = 66° C (approximately)
The final temperature of the water is 66.42 °C.
The given parameters;
The total energy released by burning the butane is calculated as follows;
Q = 10.85 x 548.9
Q = 5955.57 kcal
The final temperature of the water is calculated as follows;
[tex]Q = mc\Delta t\\\\\Delta t = \frac{Q}{mc} = \frac{5955.57}{123 \times 1} = 48.42 \ ^0C[/tex]
[tex]\Delta t = t_2 - t_1\\\\t_2 = \Delta t + t_1\\\\t_2 = 48.42 + 18\\\\t_2 = 66.42 \ ^0 C[/tex]
Thus, the final temperature of the water is 66.42 °C.
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