QuestionMay 1, 2025

A 32.0 g sample of quartz, which has a specific heat capacity of 0.730Jcdot g^-1cdot ^circ C^-1 is dropped into an insulated container containing 150,00 g of water at 15.0^circ C and a constant pressure of 1 atm.The initial temperature of the quartz is 90.8^circ C Assuming no heat is absorbed from or by the container, or the surroundings, calculate the equilibrium temperature of the water. Be sure your answer has 4 significant digits. square ^circ C

A 32.0 g sample of quartz, which has a specific heat capacity of 0.730Jcdot g^-1cdot ^circ C^-1 is dropped into an insulated container containing 150,00 g of water at 15.0^circ C and a constant pressure of 1 atm.The initial temperature of the quartz is 90.8^circ C Assuming no heat is absorbed from or by the container, or the surroundings, calculate the equilibrium temperature of the water. Be sure your answer has 4 significant digits. square ^circ C
A 32.0 g sample of quartz, which has a specific heat capacity of 0.730Jcdot g^-1cdot ^circ C^-1 is dropped into an insulated container containing 150,00 g of water at 15.0^circ C and a constant pressure of 1 atm.The initial temperature of the quartz is 90.8^circ C Assuming no heat is absorbed from or by the container, or the surroundings, calculate the equilibrium temperature of the water. Be sure your answer has 4 significant digits. square ^circ C

Solution
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Answer

T_{\text{final}} = 16.6^{\circ}C Explanation 1. Define the heat transfer equation The heat lost by quartz will be equal to the heat gained by water. We use the formula q = mc\Delta T for both substances. 2. Set up the equations for heat transfer For quartz: q_{\text{quartz}} = m_{\text{quartz}} \cdot c_{\text{quartz}} \cdot (T_{\text{final}} - T_{\text{initial, quartz}}) For water: q_{\text{water}} = m_{\text{water}} \cdot c_{\text{water}} \cdot (T_{\text{final}} - T_{\text{initial, water}}) 3. Equate the heat transfer equations Since no heat is lost to the surroundings, q_{\text{quartz}} + q_{\text{water}} = 0 4. Substitute known values and solve for T_{\text{final}} Use m_{\text{quartz}} = 32.0 \, g, c_{\text{quartz}} = 0.730 \, J \cdot g^{-1} \cdot ^{\circ}C^{-1}, T_{\text{initial, quartz}} = 90.8^{\circ}C, m_{\text{water}} = 150.0 \, g, c_{\text{water}} = 4.18 \, J \cdot g^{-1} \cdot ^{\circ}C^{-1}, T_{\text{initial, water}} = 15.0^{\circ}C Solve the equation: 32.0 \cdot 0.730 \cdot (T_{\text{final}} - 90.8) + 150.0 \cdot 4.18 \cdot (T_{\text{final}} - 15.0) = 0

Explanation

1. Define the heat transfer equation<br /> The heat lost by quartz will be equal to the heat gained by water. We use the formula $q = mc\Delta T$ for both substances.<br />2. Set up the equations for heat transfer<br /> For quartz: $q_{\text{quartz}} = m_{\text{quartz}} \cdot c_{\text{quartz}} \cdot (T_{\text{final}} - T_{\text{initial, quartz}})$<br /> For water: $q_{\text{water}} = m_{\text{water}} \cdot c_{\text{water}} \cdot (T_{\text{final}} - T_{\text{initial, water}})$<br />3. Equate the heat transfer equations<br /> Since no heat is lost to the surroundings, $q_{\text{quartz}} + q_{\text{water}} = 0$<br />4. Substitute known values and solve for $T_{\text{final}}$<br /> Use $m_{\text{quartz}} = 32.0 \, g$, $c_{\text{quartz}} = 0.730 \, J \cdot g^{-1} \cdot ^{\circ}C^{-1}$, $T_{\text{initial, quartz}} = 90.8^{\circ}C$, $m_{\text{water}} = 150.0 \, g$, $c_{\text{water}} = 4.18 \, J \cdot g^{-1} \cdot ^{\circ}C^{-1}$, $T_{\text{initial, water}} = 15.0^{\circ}C$<br /> Solve the equation: $32.0 \cdot 0.730 \cdot (T_{\text{final}} - 90.8) + 150.0 \cdot 4.18 \cdot (T_{\text{final}} - 15.0) = 0$
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