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Calculate the final temperature when 25 grams of aluminum at 80°C is placed in 100 grams of water at 25°C. The specific heat capacity of aluminum is 0.90 J/g°C and the specific heat capacity of water is 4.18 J/g°C.

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To calculate the final temperature, we can use the formula:Q_lost = Q_gainedwhere Q is the heat energy, and the heat lost by the aluminum will be equal to the heat gained by the water.For aluminum, the heat lost can be calculated using the formula:Q_Al = m_Al  c_Al   T_initial_Al - T_final where m_Al is the mass of aluminum  25 g , c_Al is the specific heat capacity of aluminum  0.90 J/gC , T_initial_Al is the initial temperature of aluminum  80C , and T_final is the final temperature.For water, the heat gained can be calculated using the formula:Q_water = m_water  c_water   T_final - T_initial_water where m_water is the mass of water  100 g , c_water is the specific heat capacity of water  4.18 J/gC , T_initial_water is the initial temperature of water  25C , and T_final is the final temperature.Now, we can set the heat lost by aluminum equal to the heat gained by water:m_Al  c_Al   T_initial_Al - T_final  = m_water  c_water   T_final - T_initial_water Plugging in the values:25 g  0.90 J/gC   80C - T_final  = 100 g  4.18 J/gC   T_final - 25C Solving for T_final:22.5   80 - T_final  = 418   T_final - 25 1800 - 22.5  T_final = 418  T_final - 104501800 + 10450 = 22.5  T_final + 418  T_final11250 = 440.5  T_finalT_final = 11250 / 440.5T_final  25.54CSo, the final temperature when 25 grams of aluminum at 80C is placed in 100 grams of water at 25C is approximately 25.54C.

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