A 440-g cylinder of brass is heated to 97.0 degree Celsius and placed in a 2 points
calorimeter containing 350 g of water at 23.0 degree Celsius. The water is
stirred, and its highest temperature is recorded as 31.0 degree Celsius.
From the thermal energy gained by the water, determine the specific heat
of brass. The specific heat of water is 4.18 J/g x C. (The heat transferred
from the brass equals the heat transferred to the water) *

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 =

C₁ = 0.40 J g⁻¹ °C⁻¹

the independent variable, or the one that is intentionally manipulated is the speed.

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