Looking for reliable answers? Westonci.ca is the ultimate Q&A platform where experts share their knowledge on various topics. Discover a wealth of knowledge from professionals across various disciplines on our user-friendly Q&A platform. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals.
Sagot :
Sure, let me guide you through the detailed steps to solve this problem.
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].
We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Westonci.ca is committed to providing accurate answers. Come back soon for more trustworthy information.