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⋅K.

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Quizplus · Accepted Answer

The specific heat capacity of the metal can be found using the principle of conservation of energy, where the heat lost by the water is equal to the heat gained by the metal. The formula to calculate the specific heat capacity ($c_m$) of the metal is derived from $q = mc\Delta T$, where $q$ is the heat transferred, $m$ is the mass, $c$ is the specific heat capacity, and $\Delta T$ is the change in temperature.Given:- Mass of water ($m_w$) = 83.0 g- Initial temperature of water ($T_{wi}$) = 78.5 °C- Final temperature of water ($T_{wf}$) = 75.9 °C- Mass of metal ($m_m$) = 66.0 g- Initial temperature of metal ($T_{mi}$) = 28.5 °C- Final temperature of metal ($T_{mf}$) = 75.9 °C (same as the final temperature of water)- Specific heat capacity of water ($c_w$) = 4.184 J/g⋅KFirst, calculate the heat lost by the water:$$q_w = m_w \cdot c_w \cdot (T_{wf} - T_{wi})$$$$q_w = 83.0 \cdot 4.184 \cdot (75.9 - 78.5)$$$$q_w = 83.0 \cdot 4.184 \cdot (-2.6)$$$$q_w = -900.5 \, \text{J} \, (\text{approximately})$$Since the heat lost by the water is equal to the heat gained by the metal:$$q_m = -q_w$$$$q_m = 900.5 \, \text{J}$$Now, calculate the specific heat capacity of the metal using the heat gained by the metal:$$q_m = m_m \cdot c_m \cdot (T_{mf} - T_{mi})$$$$900.5 = 66.0 \cdot c_m \cdot (75.9 - 28.5)$$$$c_m = \frac{900.5}{66.0 \cdot 47.4}$$$$c_m = \frac{900.5}{3128.4}$$$$c_m = 0.288 \, \text{J/g⋅K}$$Therefore, the specific heat capacity of the metal is approximately 0.29 J/g⋅K, which corresponds to choice B.

When 660 G of an Unknown Metal at 28