Question

A 28.45 g metal specimen of gold and 25.75 g metal specimen of iron, each at a temperature of 100.0 °C were dropped into 570.0 mL of water at a temperature of 17.70 °C. The molar heat capacity of iron and gold are 25.19 J mol -1 °C -1 and 25.41 J mol 1 °C -1 respectively. Assume the density of water to be 1.00g/mL, what is the final temperature of the water and pieces of metals?

Answer 1

Answer on Question #56897, Physics / Molecular Physics | Thermodynamics

A 28.45 g metal specimen of gold and 25.75 g metal specimen of iron, each at a temperature of 100.0 °C were dropped into 570.0 mL of water at a temperature of 17.70 °C. The molar heat capacity of iron and gold are 25.19 J mol⁻¹ °C⁻¹ and 25.41 J mol⁻¹ °C⁻¹ respectively. Assume the density of water to be 1.00 g/mL, what is the final temperature of the water and pieces of metals?

Solution:

This problem can be summarized thusly:

$$q_{\text{lost by gold}} + q_{\text{lost by iron}} = q_{\text{gained by water}}$$ $$c_{\text{gold}} m_{\text{gold}} (T_1 - T_x) + c_{\text{iron}} m_{\text{iron}} (T_1 - T_x) = c_{\text{water}} m_{\text{water}} (T_x - T_2)$$

The specific heat of water is = 4.187 joule/gram.

The molar mass of gold $M_{\mathrm{Au}} = 196.97 \, \mathrm{g/mol}$

The molar mass of iron $M_{\mathrm{Fe}} = 55.8450 \, \mathrm{g/mol}$

Thus,

$$c_{\text{gold}} = \frac{25.19}{196.97} = 0.128 \frac{J \, \mathrm{g/mol} \, {}^\circ\mathrm{C}}{}$$ $$c_{\text{iron}} = \frac{25.41}{55.8450} = 0.455 \frac{J \, \mathrm{g/mol} \, {}^\circ\mathrm{C}}{}$$

Here are what the several of the terms mean:

(25.75 g / 55.845 g/mol) ---> moles of Fe

(28.45 g / 196.97 g/mol) ---> moles of Au

(100.0 - x) ---> temp change of Fe and Au (they each start at 100 °C and go down to the final temperature, symbolized by 'x')

(x - 17.70) ---> temp change of water

Solve for x, which is the final temperature.

Therefore:

$$0.128 \cdot 28.45 \cdot (100 - x) + 0.455 \cdot 25.75 \cdot (100 - x) = 570 \cdot 4.187 \cdot (x - 17.70)$$ $$15.3579 \cdot (100 - x) = 2386.59 \cdot (x - 17.7)$$ $$43778.4 - 2401.95x = 0$$ $$x = 18.2262 \approx 18.2{}^{\circ}\mathrm{C}$$

Answer: $18.2{}^{\circ}\mathrm{C}$

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