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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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jacob193

0.379 J · g⁻¹ · °C⁻¹.

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.

[tex]Q = c\cdotm \cdot m\cdot \Delta T = 4.18 \times 335 \times 7.0 = 9.80\times 10^{3}\;\text{J}[/tex].

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.

[tex]Q(\text{Brass}) = Q(\text{Water})=9.80\times 10^{3}\;\text{J}[/tex].

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.

[tex]Q = c\cdot m\cdot \Delta T[/tex],

[tex]c = \dfrac{Q}{m\cdot\Delta T} =\dfrac{9.80\times 10^{3}}{410\times 63}=0.379\;\text{J}\cdot\text{g}^{-1}\cdot\textdegree{}\text{C}^{-1}[/tex].

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