To determine how many Btu 1 pound of water absorbs when it is heated from 50°F to 65°F, we can follow these steps:
1. Identify Initial and Final Temperatures:
- Initial temperature: [tex]\( 50^\circ \text{F} \)[/tex]
- Final temperature: [tex]\( 65^\circ \text{F} \)[/tex]
2. Calculate the Temperature Change ([tex]\(\Delta T\)[/tex]):
- [tex]\(\Delta T = \text{Final temperature} - \text{Initial temperature} \)[/tex]
- [tex]\(\Delta T = 65^\circ\text{F} - 50^\circ\text{F}\)[/tex]
- [tex]\(\Delta T = 15^\circ\text{F}\)[/tex]
3. Specific Heat Capacity:
- The specific heat capacity of water is a constant value of [tex]\( 1 \text{ Btu/(lb·°F)} \)[/tex].
4. Calculate the Heat Absorbed (Q):
- The formula to calculate the heat absorbed is:
[tex]\[
Q = mc\Delta T
\][/tex]
where:
- [tex]\( m \)[/tex] is the mass (1 lb in this case),
- [tex]\( c \)[/tex] is the specific heat capacity ([tex]\( 1 \text{ Btu/(lb·°F)} \)[/tex]),
- [tex]\(\Delta T\)[/tex] is the temperature change (15°F).
Given these values:
[tex]\[
Q = 1 \text{ lb} \times 1 \text{ Btu/(lb·°F)} \times 15^\circ\text{F}
\][/tex]
[tex]\[
Q = 15 \text{ Btu}
\][/tex]
Therefore, 1 pound of water heated from 50°F to 65°F absorbs 15 Btu of heat.
