Question

A 410-g cylinder of brass is heated to 95.0*C and placed in a calorimeter containing 335 g of water at 25.0*C. The water is stirred, and its highest temperature is recorded as 32.0*C. From the thermal energy gained by the water, determine the specific heat of brass. The specific heat of water is 4.18 J/g*C.

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Answer 1

0.379 J · g⁻¹ · °C⁻¹.

Step-by-step explanation

Energy change Q = c · m · ΔT.

Temperature change ΔT = Final Temperature - Initial Temperature.

The final temperature of both brass and water is 32.0 °C. However, the two differ in initial temperature.

Initial temperature

• Brass: 95.0 °C;
• Water: 25.0 °C.

How much energy did water gain in this process?

• Heat capacity c: 4.18 J · g⁻¹ · °C⁻¹;
• Mass of water m: 335 g;
• Temperature change ΔT: 32.0 - 25.0 = 7.0 °C.

`Q = c\cdotm \cdot m\cdot \Delta T = 4.18 * 335 * 7.0 = 9.80* 10^(3)\;\text{J}`.

How much energy did brass lose in this process?

The brass cylinder is cooled in a calorimeter. The calorimeter traps heat inside, such that water absorbs all the heat lost from the brass.

`Q(\text{Brass}) = Q(\text{Water})=9.80* 10^(3)\;\text{J}`.

What's the heat capacity of brass?

• Heat capacity of brass is to be found;
• Mass of the cylinder: 410 g;
• Temperature change: 95.0 - 32.0 = 63.0 °C.

`Q = c\cdot m\cdot \Delta T`,

`c = (Q)/(m\cdot\Delta T) =(9.80* 10^(3))/(410* 63)=0.379\;\text{J}\cdot\text{g}^(-1)\cdot\textdegree{}\text{C}^(-1)`.