QuestionJuly 13, 2025

A piece of metal with a mass of 32.8g is heated to 100.5^circ C and dropped into 138.2 g of water at 20.0^circ C The final temperature of the system is 30.2^circ C What is the specific heat capacity of the metal? Specific heat of water=4.184J/gC 0.391J/gC 2.56J/gC 5.29J/gC 3.5J/gC

A piece of metal with a mass of 32.8g is heated to 100.5^circ C and dropped into 138.2 g of water at 20.0^circ C The final temperature of the system is 30.2^circ C What is the specific heat capacity of the metal? Specific heat of water=4.184J/gC 0.391J/gC 2.56J/gC 5.29J/gC 3.5J/gC
A piece of metal with a mass of 32.8g is heated to 100.5^circ C and dropped into 138.2 g of water at 20.0^circ C The final temperature of the system is 30.2^circ C What is the specific heat capacity of the metal? Specific heat of water=4.184J/gC 0.391J/gC 2.56J/gC 5.29J/gC 3.5J/gC

Solution
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Answer

\( 0.391 \mathrm{~J} / \mathrm{gC} \) Explanation 1. Calculate heat absorbed by water Use q = m \cdot c \cdot \Delta T. For water: q_{\text{water}} = 138.2 \, \text{g} \times 4.184 \, \text{J/gC} \times (30.2 - 20.0) \, \text{C}. 2. Calculate heat lost by metal Heat lost by metal equals heat gained by water: q_{\text{metal}} = q_{\text{water}}. 3. Calculate specific heat capacity of metal Use c_{\text{metal}} = \frac{q_{\text{metal}}}{m_{\text{metal}} \cdot \Delta T_{\text{metal}}}. \Delta T_{\text{metal}} = 100.5 - 30.2 \, \text{C}.

Explanation

1. Calculate heat absorbed by water<br /> Use $q = m \cdot c \cdot \Delta T$. For water: $q_{\text{water}} = 138.2 \, \text{g} \times 4.184 \, \text{J/gC} \times (30.2 - 20.0) \, \text{C}$.<br />2. Calculate heat lost by metal<br /> Heat lost by metal equals heat gained by water: $q_{\text{metal}} = q_{\text{water}}$.<br />3. Calculate specific heat capacity of metal<br /> Use $c_{\text{metal}} = \frac{q_{\text{metal}}}{m_{\text{metal}} \cdot \Delta T_{\text{metal}}}$. $\Delta T_{\text{metal}} = 100.5 - 30.2 \, \text{C}$.
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