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?

23.66°C

Explanation:

We are given;

Mass of Pb as 20.0 g Amount of heat transferred is 4950 Joules Mass of water is 50.0 g

We are required to calculate the temperature change

Note that the quantity of heat, Q absorbed is given by;

Q = mass * specific heat capacity * Change in temperature

In this case;

Specific heat capacity, c of water = 4.184 J/g°C

But, Q = mcΔT

Amount of heat absorbed by water is 4950 J

Rearranging the formula;

ΔT = Q : mc

= 4950 J : (50.0 g * 4.184 J/g°C)

= 23.66°C

Thus, the temperature of water will change by 23.66°C