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Answer :
To calculate the heat [tex]\( q \)[/tex] that would be given off when a lead bar is cooled, you can use the formula:
[tex]\[ q = m \times c \times \Delta T \][/tex]
where:
- [tex]\( q \)[/tex] is the heat exchanged, in joules (J).
- [tex]\( m \)[/tex] is the mass of the substance, in grams (g).
- [tex]\( c \)[/tex] is the specific heat capacity, in joules per gram per degree Celsius (J/g°C).
- [tex]\( \Delta T \)[/tex] is the change in temperature, in degrees Celsius (°C).
Let's break it down step-by-step:
1. Identify the Given Values:
- Mass of the lead bar, [tex]\( m = 458 \)[/tex] g.
- Initial temperature, [tex]\( T_{\text{initial}} = 97.5 \)[/tex] °C.
- Final temperature, [tex]\( T_{\text{final}} = 47.5 \)[/tex] °C.
- Specific heat capacity of lead, [tex]\( c = 0.128 \)[/tex] J/g°C.
2. Calculate the Change in Temperature ([tex]\( \Delta T \)[/tex]):
[tex]\[
\Delta T = T_{\text{final}} - T_{\text{initial}} = 47.5\,\text{°C} - 97.5\,\text{°C} = -50.0\,\text{°C}
\][/tex]
3. Calculate the Heat Given Off ([tex]\( q \)[/tex]):
[tex]\[
q = 458\,\text{g} \times 0.128\,\text{J/g°C} \times (-50.0\,\text{°C})
\][/tex]
[tex]\[
q = -2931.2\,\text{J}
\][/tex]
The negative sign indicates that the heat [tex]\( q \)[/tex] is given off by the lead bar as it cools down. Therefore, the lead bar gives off 2931.2 joules of heat as it cools from 97.5 °C to 47.5 °C.
[tex]\[ q = m \times c \times \Delta T \][/tex]
where:
- [tex]\( q \)[/tex] is the heat exchanged, in joules (J).
- [tex]\( m \)[/tex] is the mass of the substance, in grams (g).
- [tex]\( c \)[/tex] is the specific heat capacity, in joules per gram per degree Celsius (J/g°C).
- [tex]\( \Delta T \)[/tex] is the change in temperature, in degrees Celsius (°C).
Let's break it down step-by-step:
1. Identify the Given Values:
- Mass of the lead bar, [tex]\( m = 458 \)[/tex] g.
- Initial temperature, [tex]\( T_{\text{initial}} = 97.5 \)[/tex] °C.
- Final temperature, [tex]\( T_{\text{final}} = 47.5 \)[/tex] °C.
- Specific heat capacity of lead, [tex]\( c = 0.128 \)[/tex] J/g°C.
2. Calculate the Change in Temperature ([tex]\( \Delta T \)[/tex]):
[tex]\[
\Delta T = T_{\text{final}} - T_{\text{initial}} = 47.5\,\text{°C} - 97.5\,\text{°C} = -50.0\,\text{°C}
\][/tex]
3. Calculate the Heat Given Off ([tex]\( q \)[/tex]):
[tex]\[
q = 458\,\text{g} \times 0.128\,\text{J/g°C} \times (-50.0\,\text{°C})
\][/tex]
[tex]\[
q = -2931.2\,\text{J}
\][/tex]
The negative sign indicates that the heat [tex]\( q \)[/tex] is given off by the lead bar as it cools down. Therefore, the lead bar gives off 2931.2 joules of heat as it cools from 97.5 °C to 47.5 °C.
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