How much heat energy was added to water that started at 15°C and reached 80°C?

How much heat energy was added to water that started at 15°C and reached 80°C?

• A. 100 Btu
• B. 120 Btu
• C. 117 Btu
• D. 110 Btu

Answer: (C)

To determine how much heat energy was added to the water, you can use the formula for heat transfer, which is: \[ Q = m \cdot c \cdot \Delta T \] Where: - \( Q \) is the heat added (in Btu), - \( m \) is the mass of the water (in pounds), - \( c \) is the specific heat capacity of water (which is typically about 1 Btu/lb·°F), - \( \Delta T \) is the change in temperature (in °F). In this scenario, the water starts at 15°C and ends at 80°C. First, we need to convert these temperatures to Fahrenheit because the specific heat is given in Btu/lb·°F. The conversion formula from Celsius to Fahrenheit is: \[ °F = (°C \times \frac{9}{5}) + 32 \] Calculating for the two temperatures: - 15°C = (15 × 9/5) + 32 = 59°F - 80°C = (80 × 9/5) + 32 = 176°F Now, we can find the change in temperature (\( \Delta

To determine how much heat energy was added to the water, you can use the formula for heat transfer, which is:

[ Q = m \cdot c \cdot \Delta T ]

Where:

• ( Q ) is the heat added (in Btu),

• ( m ) is the mass of the water (in pounds),

• ( c ) is the specific heat capacity of water (which is typically about 1 Btu/lb·°F),

• ( \Delta T ) is the change in temperature (in °F).

In this scenario, the water starts at 15°C and ends at 80°C. First, we need to convert these temperatures to Fahrenheit because the specific heat is given in Btu/lb·°F. The conversion formula from Celsius to Fahrenheit is:

[ °F = (°C \times \frac{9}{5}) + 32 ]

Calculating for the two temperatures:

• 15°C = (15 × 9/5) + 32 = 59°F

• 80°C = (80 × 9/5) + 32 = 176°F

Now, we can find the change in temperature (( \Delta