Question

If 20.0 g of Pb transfers 4950 J of energy into 50.0 g of water, how much would the temperature of the water change?​ By how much would the metal cool?

Answer 1

To find the temperature change of water, we'll use the formula:

q = mcΔT, where q is the heat energy transferred, m is the mass, c is the specific heat capacity, and ΔT is the temperature change.

For water, the specific heat capacity is 4.18 J/g°C.

First, let's find the temperature change of water:

4950 J = 50.0 g × 4.18 J/g°C × ΔT

ΔT = 4950 J / (50.0 g × 4.18 J/g°C) ≈ 23.68°C

Now, let's find the temperature change of the Pb metal. Since the energy transferred to the water is the same as the energy lost by the metal, we can use the same formula:

4950 J = 20.0 g × c_Pb × ΔT_metal

The specific heat capacity of Pb is 0.128 J/g°C.

4950 J = 20.0 g × 0.128 J/g°C × ΔT_metal

ΔT_metal = 4950 J / (20.0 g × 0.128 J/g°C) ≈ 193.36°C

The temperature of the water would change by approximately 23.68°C, and the metal would cool by approximately 193.36°C.