Question

The specific heat capacity of lead is 0.13 J/g-K. How much heat (in J) is required to raise the temperature of 15 g of lead from 22 °C to 37 °C? a. 5.8 × 10-4 J b. 0.13 J c. 29 J d. 2.0 J e. -0.13 J

Answer 1

c. 29 J

Step-by-step explanation:

Step 1: Given data

• Specific heat capacity of Pb (c): 0.13 J/g.K (= 0.13 J/g.°C)

• Mass of Pb (m): 15 g

• Initial temperature: 22 °C

• Final temperature: 37 °C

Step 2: Calculate the temperature change

ΔT = 37 °C - 22 °C = 15 °C

Step 3: Calculate the heat (Q) required to raise the temperature of the lead piece

We will use the following expression.

Q = c × m × ΔT

Q = 0.13 J/g.°C × 15 g × 15 °C = 29 J