To 209 g of water contained in a 95.4 g glass beaker at 20.9 °C, was added 523.5 g of aluminum metal that was heated to 95.6 °C. If the specific heat capacity of water, glass and aluminum are 4.15 J/g-°C, 0.89 J/g-°C, and 0.31 J/g-°C respectively, calculate the final temperature (°C) of the aluminum-glass-water mixture. Submit your answer with no units and to two decimal places.

 

31.78

Step-by-step explanation

• heat lost = heat gained
• heat lost by aluminum = heat gained by water + heat gained by glass
• Q=mCdT
• thus
• mCdT_aluminium=mCdT_water+ mCdT_glass
• 523.5g x 0.31 J/g-°C x (95.6-T) = [209g x 4.15 J/g-°C (T-20.9)] + [95.4g x 0.89 J/g-°C x (T-20.9)
• 15515.9 -162.3T = [867.35T -18127.62] + [84.9T - 1774.54]
• 15515.9 -162.3T = 867.35T -18127.62 + 84.9T - 1774.54
• we collect like terms
• 15515.9 -162.3T =952.25T - 19902.16
• 35418.06=1114.55T
• T=31.78⁰C
• ANS: 31.78 (with no units as requested)