A 5 g sample of lead (specific heat 0.129 / g˚C) is heated, then put in a calorimeter with 50 mL of water (specific heat 4.184 J/g˚C). The water temperature rises from 20∘C to 22∘C. If the lead sample's final temperature was also 22∘C, what was the initial temperature of the lead sample?

670.68°C

Explanation:

Given that:

volume of water = 50 ml but 1 g = 1 ml. Therefore the mass of water (m) = 50 ml * 1 g / ml = 50 g

specific heat (C) = 4.184 J/g˚C

Initial temperature = 20°C, final temperature = 22°C. Therefore the temperature change ΔT = final temperature - initial temperature = 22 - 20 = 2°C

The quantity of heat (Q) used to raise the temperature of a body is given by the equation:

Q = mCΔT

Substituting values:

Q = 50 g * 4.184 J/g˚C * 2°C = 418.4 J

Since the mass of lead = 5 g and specific heat = 0.129 J/g˚C. The heat used to raise the temperature of water is the same heat used to raise the temperature of lead.

-Q = mCΔT

-418.4 J = 5 g * 0.129 J/g˚C * ΔT

ΔT = - 418.4 J / (5 g * 0.129 J/g˚C) = - 648.68°C

temperature change ΔT = final temperature - initial temperature

- 648.68°C = 22°C - Initial Temperature

Initial Temperature = 22 + 648.68 = 670.68°C