QuestionJune 8, 2025

Question 6 (10 points) Calculate the final temperature when 23.4 grams of water at 75 degrees C is mixed with 2.45 grams of iron at 2.6 degrees C. The specific heat of water is 4.184J/gC and the specific heat of iron is 0.45J/gC Show your work

Question 6 (10 points) Calculate the final temperature when 23.4 grams of water at 75 degrees C is mixed with 2.45 grams of iron at 2.6 degrees C. The specific heat of water is 4.184J/gC and the specific heat of iron is 0.45J/gC Show your work
Question 6 (10 points) Calculate the final temperature when 23.4 grams of water at 75 degrees C is mixed with 2.45 grams of iron at 2.6 degrees C. The specific heat of water is 4.184J/gC and the specific heat of iron is 0.45J/gC Show your work

Solution
4.4(210 votes)

Answer

T_f \approx 74.19^\circ C Explanation 1. Define Heat Transfer Equation Use the formula for heat transfer: **q = mc\Delta T**. The heat lost by water equals the heat gained by iron. 2. Set Up Equations for Water and Iron For water: q_{\text{water}} = m_{\text{water}} \cdot c_{\text{water}} \cdot (T_f - T_{\text{initial, water}}). For iron: q_{\text{iron}} = m_{\text{iron}} \cdot c_{\text{iron}} \cdot (T_f - T_{\text{initial, iron}}). 3. Apply Conservation of Energy Since no heat is lost to surroundings, q_{\text{water}} + q_{\text{iron}} = 0. 4. Substitute Values and Solve 23.4 \cdot 4.184 \cdot (T_f - 75) + 2.45 \cdot 0.45 \cdot (T_f - 2.6) = 0. 5. Simplify and Solve for T_f 97.9056(T_f - 75) + 1.1025(T_f - 2.6) = 0. 97.9056T_f - 7342.92 + 1.1025T_f - 2.8665 = 0. 99.0081T_f = 7345.7865. T_f = \frac{7345.7865}{99.0081}.

Explanation

1. Define Heat Transfer Equation<br /> Use the formula for heat transfer: **$q = mc\Delta T$**. The heat lost by water equals the heat gained by iron.<br /><br />2. Set Up Equations for Water and Iron<br /> For water: $q_{\text{water}} = m_{\text{water}} \cdot c_{\text{water}} \cdot (T_f - T_{\text{initial, water}})$.<br /> For iron: $q_{\text{iron}} = m_{\text{iron}} \cdot c_{\text{iron}} \cdot (T_f - T_{\text{initial, iron}})$.<br /><br />3. Apply Conservation of Energy<br /> Since no heat is lost to surroundings, $q_{\text{water}} + q_{\text{iron}} = 0$.<br /><br />4. Substitute Values and Solve<br /> $23.4 \cdot 4.184 \cdot (T_f - 75) + 2.45 \cdot 0.45 \cdot (T_f - 2.6) = 0$.<br /><br />5. Simplify and Solve for $T_f$<br /> $97.9056(T_f - 75) + 1.1025(T_f - 2.6) = 0$.<br /> $97.9056T_f - 7342.92 + 1.1025T_f - 2.8665 = 0$.<br /> $99.0081T_f = 7345.7865$.<br /> $T_f = \frac{7345.7865}{99.0081}$.
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