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1. Given Data:
- Specific heat capacity of lead ([tex]\(c\)[/tex]): [tex]\(0.129 \, \text{J/g-K}\)[/tex]
- Mass of the lead ([tex]\(m\)[/tex]): [tex]\(382 \, \text{g}\)[/tex]
- Initial temperature ([tex]\(T_i\)[/tex]): [tex]\(22.50 \, \text{°C}\)[/tex]
- Final temperature ([tex]\(T_f\)[/tex]): [tex]\(37.20 \, \text{°C}\)[/tex]
2. Determine the Temperature Change ([tex]\(\Delta T\)[/tex]):
[tex]\[ \Delta T = T_f - T_i \][/tex]
[tex]\[ \Delta T = 37.20 \, \text{°C} - 22.50 \, \text{°C} = 14.70 \, \text{°C} \][/tex]
3. Calculate the Heat Required ([tex]\(Q\)[/tex]):
The formula to calculate the heat required is:
[tex]\[ Q = mc\Delta T \][/tex]
Substituting the values we have:
[tex]\[ Q = 382 \, \text{g} \times 0.129 \, \text{J/g-K} \times 14.70 \, \text{K} \][/tex]
4. Perform the Multiplication:
[tex]\[ Q = 382 \times 0.129 \times 14.70 = 724.39 \, \text{J} \][/tex]
Therefore, the number of joules of heat required to raise the temperature of 382 g of lead from 22.50°C to 37.20°C is approximately [tex]\(724.39 \, \text{J}\)[/tex].
1. Given Data:
- Specific heat capacity of lead ([tex]\(c\)[/tex]): [tex]\(0.129 \, \text{J/g-K}\)[/tex]
- Mass of the lead ([tex]\(m\)[/tex]): [tex]\(382 \, \text{g}\)[/tex]
- Initial temperature ([tex]\(T_i\)[/tex]): [tex]\(22.50 \, \text{°C}\)[/tex]
- Final temperature ([tex]\(T_f\)[/tex]): [tex]\(37.20 \, \text{°C}\)[/tex]
2. Determine the Temperature Change ([tex]\(\Delta T\)[/tex]):
[tex]\[ \Delta T = T_f - T_i \][/tex]
[tex]\[ \Delta T = 37.20 \, \text{°C} - 22.50 \, \text{°C} = 14.70 \, \text{°C} \][/tex]
3. Calculate the Heat Required ([tex]\(Q\)[/tex]):
The formula to calculate the heat required is:
[tex]\[ Q = mc\Delta T \][/tex]
Substituting the values we have:
[tex]\[ Q = 382 \, \text{g} \times 0.129 \, \text{J/g-K} \times 14.70 \, \text{K} \][/tex]
4. Perform the Multiplication:
[tex]\[ Q = 382 \times 0.129 \times 14.70 = 724.39 \, \text{J} \][/tex]
Therefore, the number of joules of heat required to raise the temperature of 382 g of lead from 22.50°C to 37.20°C is approximately [tex]\(724.39 \, \text{J}\)[/tex].
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