QuestionJuly 16, 2025

A 100.0 g piece of iron (c=0.449J/g^circ C) at 95.0^circ C is placed in 200.0 g of water (c=4.184J/g^circ C) initially at 20.0^circ C What is the final temperature of the system? 23.6^circ C 57.5^circ C 27.2^circ C 31.4^circ C

A 100.0 g piece of iron (c=0.449J/g^circ C) at 95.0^circ C is placed in 200.0 g of water (c=4.184J/g^circ C) initially at 20.0^circ C What is the final temperature of the system? 23.6^circ C 57.5^circ C 27.2^circ C 31.4^circ C
A 100.0 g piece of iron (c=0.449J/g^circ C) at 95.0^circ C is placed in 200.0 g of water (c=4.184J/g^circ C) initially at 20.0^circ C What is the final temperature of the system? 23.6^circ C 57.5^circ C 27.2^circ C 31.4^circ C

Solution
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Answer

23.6^{\circ}C Explanation 1. Define heat transfer equation Use q_{\text{iron}} = -q_{\text{water}} where q = mc\Delta T. 2. Set up equations for iron and water For iron: 100.0 \times 0.449 \times (T_f - 95.0); For water: 200.0 \times 4.184 \times (T_f - 20.0). 3. Equate and solve for T_f 100.0 \times 0.449 \times (T_f - 95.0) = -200.0 \times 4.184 \times (T_f - 20.0). Simplify and solve: 44.9(T_f - 95.0) = -836.8(T_f - 20.0). 44.9T_f - 4265.5 = -836.8T_f + 16736. 881.7T_f = 20991.5. T_f = \frac{20991.5}{881.7} \approx 23.8^{\circ}C.

Explanation

1. Define heat transfer equation<br /> Use $q_{\text{iron}} = -q_{\text{water}}$ where $q = mc\Delta T$.<br />2. Set up equations for iron and water<br /> For iron: $100.0 \times 0.449 \times (T_f - 95.0)$; For water: $200.0 \times 4.184 \times (T_f - 20.0)$.<br />3. Equate and solve for $T_f$<br /> $100.0 \times 0.449 \times (T_f - 95.0) = -200.0 \times 4.184 \times (T_f - 20.0)$.<br /> Simplify and solve: $44.9(T_f - 95.0) = -836.8(T_f - 20.0)$.<br /> $44.9T_f - 4265.5 = -836.8T_f + 16736$.<br /> $881.7T_f = 20991.5$.<br /> $T_f = \frac{20991.5}{881.7} \approx 23.8^{\circ}C$.
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