QuestionDecember 17, 2025

2. Suppose that hot coffee has a temperature of 90^circ C in the coffee pot.You pour a serving of volume of 0.10 L into a ceramic coffee mug (specific heat 940J/kg^circ C) that has a mass of 0.30 kg and is initially at 20^circ C . What is the temperature of your coffee when the coffee and mug reach thermal equilibrium?

2. Suppose that hot coffee has a temperature of 90^circ C in the coffee pot.You pour a serving of volume of 0.10 L into a ceramic coffee mug (specific heat 940J/kg^circ C) that has a mass of 0.30 kg and is initially at 20^circ C . What is the temperature of your coffee when the coffee and mug reach thermal equilibrium?
2. Suppose that hot coffee has a temperature of 90^circ C in the coffee pot.You pour a serving of volume of 0.10 L into a ceramic coffee mug (specific heat 940J/kg^circ C) that has a mass of 0.30 kg and is initially at 20^circ C . What is the temperature of your coffee when the coffee and mug reach thermal equilibrium?

Solution
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Answer

61.8^\circ\text{C} Explanation 1. Set up energy conservation equation At equilibrium, heat lost by coffee = heat gained by mug: m_c c_c (T_f - T_{c,i}) + m_m c_m (T_f - T_{m,i}) = 0. 2. Assign variables and values m_c = 0.10\,\text{kg} (assuming density of water), c_c = 4186\,\text{J/kg}^\circ\text{C}, T_{c,i} = 90^\circ\text{C}; m_m = 0.30\,\text{kg}, c_m = 940\,\text{J/kg}^\circ\text{C}, T_{m,i} = 20^\circ\text{C}. 3. Substitute values into equation 0.10 \times 4186 \times (T_f - 90) + 0.30 \times 940 \times (T_f - 20) = 0 4. Expand and combine like terms 418.6(T_f - 90) + 282(T_f - 20) = 0 418.6T_f - 37674 + 282T_f - 5640 = 0 (418.6 + 282)T_f = 37674 + 5640 700.6T_f = 43314 5. Solve for final temperature T_f = \frac{43314}{700.6} \approx 61.8^\circ\text{C}

Explanation

1. Set up energy conservation equation<br /> At equilibrium, heat lost by coffee = heat gained by mug: $m_c c_c (T_f - T_{c,i}) + m_m c_m (T_f - T_{m,i}) = 0$.<br />2. Assign variables and values<br /> $m_c = 0.10\,\text{kg}$ (assuming density of water), $c_c = 4186\,\text{J/kg}^\circ\text{C}$, $T_{c,i} = 90^\circ\text{C}$; $m_m = 0.30\,\text{kg}$, $c_m = 940\,\text{J/kg}^\circ\text{C}$, $T_{m,i} = 20^\circ\text{C}$.<br />3. Substitute values into equation<br /> $0.10 \times 4186 \times (T_f - 90) + 0.30 \times 940 \times (T_f - 20) = 0$<br />4. Expand and combine like terms<br /> $418.6(T_f - 90) + 282(T_f - 20) = 0$<br /> $418.6T_f - 37674 + 282T_f - 5640 = 0$<br /> $(418.6 + 282)T_f = 37674 + 5640$<br /> $700.6T_f = 43314$<br />5. Solve for final temperature<br /> $T_f = \frac{43314}{700.6} \approx 61.8^\circ\text{C}$
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