Question

Suppose you heat a metal object with a mass of 34.5 g to 95.5 °C and transfer it to a calorimeter containing 100.0 g of water at 17.5 °C. The water and metal reach a final temperature of 24.9 °C.

Metal in a covered cup with a thermometer and heat indicated as leaving the metal.

What is the specific heat of the metal in J/g⋅∘C

Answer 1

The specific heat of the metal can be calculated using the formula:

q = m × c × ΔT

where q is the heat transferred, m is the mass of the metal, c is the specific heat of the metal, and ΔT is the change in temperature.

First, we need to calculate the heat transferred from the metal to the water:

q = m × c × ΔT
q = (34.5 g) × c × (95.5 °C - 24.9 °C)
q = 224,085 J

Next, we can calculate the heat absorbed by the water:

q = m × c × ΔT
q = (100.0 g) × (4.184 J/g⋅∘C) × (24.9 °C - 17.5 °C)
q = 3,073 J

Since the heat transferred from the metal to the water is equal to the heat absorbed by the water, we can set the two equations equal to each other and solve for c:

m × c × ΔT = m × c × ΔT
(34.5 g) × c × (95.5 °C - 24.9 °C) = (100.0 g) × (4.184 J/g⋅∘C) × (24.9 °C - 17.5 °C)
c = 0.385 J/g⋅∘C

Therefore, the specific heat of the metal is 0.385 J/g⋅∘C.