In order to heat 5 pounds of water from 45 degrees to 55 degrees, that water would have to absorb:

To calculate the energy required to heat water, we use the formula:

\[
Q = mc\Delta T
\]

Where:
- \(Q\) is the heat energy absorbed (in calories or joules),
- \(m\) is the mass of the water (in grams),
- \(c\) is the specific heat capacity of water (1 calorie/gram°C or 4.184 joules/gram°C),
- \(\Delta T\) is the change in temperature.

### Given:
- Mass \(m = 5\) pounds = \(5 \times 453.592 \, \text{g} = 2267.96 \, \text{g}\),
- Change in temperature \(\Delta T = 55^\circ F - 45^\circ F = 10^\circ F = \frac{10}{1.8} \, \text{C} \approx 5.56^\circ C\),
- Specific heat \(c = 1 \, \text{cal/g°C}\) (or \(4.184 \, \text{J/g°C}\)).

### Calculating:

\[
Q = 2267.96 \, \text{g} \times 1 \, \text{cal/g°C} \times 5.56 \, \text{°C}
\]
Let's calculate this.The water would need to absorb approximately 12,600 calories of heat energy to increase its temperature from 45°F to 55°F.

Would you like more details or have any questions?

Here are five related questions for deeper understanding:
1. How do you convert between calories and joules?
2. What factors affect the specific heat of a substance?
3. How does the energy requirement change if the mass of the water is doubled?
4. What is the energy required if the temperature change is 20°F instead of 10°F?
5. Can the specific heat capacity of water change with temperature?

**Tip:** Always make sure to use consistent units when calculating heat energy, such as converting pounds to grams or Fahrenheit to Celsius.

Learn how to calculate the heat energy required to raise the temperature of 5 pounds of water from 45°F to 55°F. This problem covers thermodynamic concepts, heat transfer formulas, and unit conversions between Fahrenheit and Celsius. Suitable for high school students in Grades 9-12.

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