A 170.0-g sample of metal at 83.00°C is added to 170.0 g of H₂O(l)at 15.00°C in an insulated container.The temperature rises to 18.16°C.Neglecting the heat capacity of the container,what is the specific heat capacity of the metal? The specific heat capacity of H₂O(l)is 4.18 J/(g·°C).

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Quizplus · Accepted Answer

The specific heat capacity of the metal can be found using the principle of conservation of energy, which states that the heat lost by the metal will be equal to the heat gained by the water. 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.For the metal:$$q_m = m_m \cdot c_m \cdot (T_f - T_i)$$For the water (H₂O(l) is assumed to be water with a typo):$$q_w = m_w \cdot c_w \cdot (T_f - T_i)$$Given that the heat lost by the metal is equal to the heat gained by the water, and neglecting the heat capacity of the container:$$m_m \cdot c_m \cdot (T_f - T_i)_m = m_w \cdot c_w \cdot (T_f - T_i)_w$$Given:- $m_m = m_w = 170.0 \, g$- $c_w = 4.18 \, J/(g·°C)$- $T_{i,m} = 83.00°C$, $T_{i,w} = 15.00°C$, $T_f = 18.16°C$Substituting the given values:$$170.0 \cdot c_m \cdot (18.16 - 83.00) = 170.0 \cdot 4.18 \cdot (18.16 - 15.00)$$Solving for $c_m$:$$c_m = \frac{170.0 \cdot 4.18 \cdot (18.16 - 15.00)}{170.0 \cdot (18.16 - 83.00)}$$$$c_m = \frac{4.18 \cdot 3.16}{-64.84}$$$$c_m \approx 0.204 \, J/(g·°C)$$Therefore, the specific heat capacity of the metal is approximately 0.204 J/(g·°C).

A 1700-G Sample of Metal at 83