Question

A 440-g cylinder of brass is heated to 97.0 degree Celsius and placed in a 2 points

Answer 1

Answer:

Specific heat of brass is 0.40 J g⁻¹ °C⁻¹ .

Explanation:

Given :

Mass of brass, m₁ = 440 g

Temperature of brass, T₁ = 97° C

Mass of water, m₂ = 350 g

Temperature of water, T₂ = 23° C

Specific heat of water, C₂ = 4.18 J g⁻¹ °C⁻¹

Equilibrium temperature, T = 31° C

Let C₁ be the specific heat of brass.

Heat loss by brass = Heat gain by water

m₁ x C₁ x ( T₁ -T ) = m₂ x C₂ x ( T - T₁ )

Substitute the suitable values in above equation.

440 x C₁ x (97 - 31) = 350 x 4.18 x (31 - 23)

C₁ = $\frac{11704}{29040}$

C₁ = 0.40 J g⁻¹ °C⁻¹