Question

A piece of metal with a mass of 15.3 grams has a temperature of 50.0 ºC. When the metal is placed in 80.2 grams of water at 21.0ºC, the temperature rises by 4.3 ºC. What is the specific heat capacity of the metal?

Answer 1

Given:

Mass of metal, m₁ = 15.3 gm

Temperature of Metal, T₁ = 50°C

Mass of water, m₂ = 80.2 gm

Temperature of water, T₂ = 21°C

Change in temperature of water, ΔT = 4.3°C

To Find:

Specific heat capacity of metal

Calculation:

- Let the specific heat capacity of metal be c₁ and specific heat capacity of  water c₂ be 4.186 × 10³ J/kgK

- Change in temp of metal ΔT' = 50 - 25.3 = 24.7°C

- Since we know that

Heat lost by the piece of metal = Heat gained by the water

⇒ m₁ × c₁ × ΔT' = m₂ × c₂ × ΔT

⇒ 15.3 × c₁ × 24.7 = 80.2 × 4.186 × 10³ × 4.3

⇒ c₁ =  1443.584 × 10³ / 377.91

⇒ c₁ =  3.8199 × 10³ J/kgK

⇒ c₁ =  3.820 × 10³ J/kgK