When 66.0 g of an unknown metal at 28.5°C is placed in 83.0 g H₂O at 78.5°C,the water temperature decreases to 75.9°C.What is the specific heat capacity of the metal? The specific heat capacity of water is 4.184 J/g⋅°C.

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Quizplus · Accepted Answer

To find the specific heat capacity of the metal, we use the principle of conservation of energy, where the heat lost by the water is equal to the heat gained by the metal. The formula for heat (q) is q = mcΔT, where m is mass, c is specific heat capacity, and ΔT is the change in temperature.For water:- Mass (m) = 83.0 g- Specific heat capacity (c) = 4.184 J/g°C- Change in temperature (ΔT) = 75.9°C - 78.5°C = -2.6°CFor the metal:- Mass (m) = 66.0 g- Specific heat capacity (c) = unknown- Change in temperature (ΔT) = 75.9°C - 28.5°C = 47.4°CSetting the heat gained by the metal equal to the heat lost by the water:$$q_{metal} = q_{water}$$$$m_{metal}c_{metal}\Delta T_{metal} = m_{water}c_{water}\Delta T_{water}$$$$66.0 \times c_{metal} \times 47.4 = 83.0 \times 4.184 \times (-2.6)$$Solving for $c_{metal}$:$$c_{metal} = \frac{83.0 \times 4.184 \times (-2.6)}{66.0 \times 47.4}$$$$c_{metal} \approx 0.29 \, \text{J/g°C}$$Therefore, the specific heat capacity of the metal is approximately 0.29 J/g°C.

When 660 G of an Unknown Metal at 28