A 50.0 g glass container contains $200 \mathrm{~g}$ of water at $20.0^{\circ} \mathrm{C}$. A $20.0 \mathrm{~g}$ piece of an alloy at $100{ }^{\circ} \mathrm{C}$ is added to the water in the container. The final temperature of the mixture is $24^{\circ} \mathrm{C}$. What is the specific heat of the alloy in cal/g ${ }^{\circ} \mathrm{C}$ ? (specific heat of water $=1.00 \mathrm{cal} / \mathrm{g}{ }^{\circ} \mathrm{C}$, specific heat of glass $=0.200 \mathrm{cal} / \mathrm{g}{ }^{\circ} \mathrm{C}$ )

Question

Quizplus · Accepted Answer

The specific heat of the alloy can be calculated using the principle of conservation of energy, where the heat lost by the alloy equals the heat gained by the water and the glass container. The formula for heat is $q = mc\Delta T$, where $m$ is mass, $c$ is specific heat, and $\Delta T$ is the change in temperature.For the alloy:$$q_{\text{alloy}} = m_{\text{alloy}} \times c_{\text{alloy}} \times (T_{\text{final}} - T_{\text{initial}})$$For the water:$$q_{\text{water}} = m_{\text{water}} \times c_{\text{water}} \times (T_{\text{final}} - T_{\text{initial}})$$For the glass:$$q_{\text{glass}} = m_{\text{glass}} \times c_{\text{glass}} \times (T_{\text{final}} - T_{\text{initial}})$$Given:- $m_{\text{alloy}} = 20.0 \, \text{g}$- $m_{\text{water}} = 200.0 \, \text{g}$- $m_{\text{glass}} = 50.0 \, \text{g}$- $c_{\text{water}} = 1.00 \, \text{cal/g}^\circ\text{C}$- $c_{\text{glass}} = 0.200 \, \text{cal/g}^\circ\text{C}$- $T_{\text{final}} = 24.0^\circ\text{C}$- $T_{\text{initial, alloy}} = 100.0^\circ\text{C}$- $T_{\text{initial, water and glass}} = 20.0^\circ\text{C}$The heat lost by the alloy is equal to the heat gained by the water and the glass:$$m_{\text{alloy}} \times c_{\text{alloy}} \times (24 - 100) = m_{\text{water}} \times c_{\text{water}} \times (24 - 20) + m_{\text{glass}} \times c_{\text{glass}} \times (24 - 20)$$Solving for $c_{\text{alloy}}$:$$20 \times c_{\text{alloy}} \times (-76) = 200 \times 1 \times 4 + 50 \times 0.200 \times 4$$$$c_{\text{alloy}} = \frac{200 \times 4 + 50 \times 0.200 \times 4}{20 \times (-76)}$$$$c_{\text{alloy}} = \frac{800 + 40}{-1520}$$$$c_{\text{alloy}} = \frac{840}{-1520} = -0.553 \, \text{cal/g}^\circ\text{C}$$Since specific heat cannot be negative, the positive value of $0.553 \, \text{cal/g}^\circ\text{C}$ is the correct magnitude for the specific heat of the alloy.

A 500 G Glass Container Contains 200 g200 \mathrm{~g}200 g Of Water At 20.0∘C20.0^{\circ} \mathrm{C}20.0∘C

A 500 G Glass Container Contains $200 \mathrm{~g}$ Of Water At $20.0^{\circ} \mathrm{C}$