Welcome to Westonci.ca, the Q&A platform where your questions are met with detailed answers from experienced experts. Experience the convenience of getting reliable answers to your questions from a vast network of knowledgeable experts. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform.

The specific heat of lead is 0.129 J/g-K. How many joules of heat are required to raise the temperature of 382 g of lead from 22.50°C to 37.20°C?

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].