Question

The BTU (British thermal unit) is the unit of energy most commonly used in the United States. One joule = \(9.48 \times 10^{-4} \mathrm{BTU}\). What is the specific heat of water in \(\mathrm{BTU} /\) lb \(\cdot{ }^{\circ} \mathrm{F} ?\) (Specific heat of water is \(\left.4.18 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C} .\right)\)

Short answer

Question: Convert the specific heat of water from 4.18 J/g·°C to BTU/lb·°F. Answer: The specific heat of water in BTU/lb·°F is approximately \(4.85 \times 10^{-6} \mathrm{BTU/lb} \cdot{ }^{\circ} \mathrm{F}\).

Step-by-step solution

Convert J to BTU

Multiply the given specific heat of water by the conversion factor of J to BTU: \(4.18 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C} \times 9.48 \times 10^{-4} \mathrm{BTU/J} = 3.96184 \times 10^{-3} \mathrm{BTU/g} \cdot{ }^{\circ} \mathrm{C}\)

Convert g to lb

Use the conversion factor of 1 lb = 453.592 g to convert g to lb: \(3.96184 \times 10^{-3} \mathrm{BTU/g} \cdot{ }^{\circ} \mathrm{C} \times \frac{1 \mathrm{lb}}{453.592 \mathrm{g}} = 8.73307 \times 10^{-6} \mathrm{BTU/lb} \cdot{ }^{\circ} \mathrm{C}\)

Convert °C to °F

Since 1 °C = 1.8 °F, we need to divide the specific heat by 1.8 to convert it to BTU/lb·°F: \(8.73307 \times 10^{-6} \mathrm{BTU/lb} \cdot{ }^{\circ} \mathrm{C} \times \frac{1}{1.8} = 4.85170 \times 10^{-6} \mathrm{BTU/lb} \cdot{ }^{\circ} \mathrm{F}\) Therefore, the specific heat of water in BTU/lb·°F is approximately \(4.85 \times 10^{-6} \mathrm{BTU/lb} \cdot{ }^{\circ} \mathrm{F}\).