A chemistry professor has a cup of coffee containing 50.0mL of room temperature coffee at
25.0C. The professor also has a new pot of hot coffee at temperature of 96.0C. What volume of hot
coffee will the professor need to add to his cold coffee to reach the ideal drinking temperature of
82.0C? Assume that no heat is lost to the coffee cup or to the environment. Also assume that coffee
has the same density (1.0g/mL) and heat capacity (4.184 J/gC) as water.
3 answers
Use the same formula as in the warm/hot water problem but this time you know Tf and Ti and you want to solve for mass hot coffee.
How do you find Q for this example?
I did
50mL x 4.184 x (82-25) = Heat gained = 11924J
Heat gained 11924J= Heat lost 11924J
Heat lost = 11924J = Mass x 4.184 x (82-96)
11924J / (4.184 x -14) = Mass
Mass = 203.56 = 204 mL (with proper significant figures)
Is this right?
50mL x 4.184 x (82-25) = Heat gained = 11924J
Heat gained 11924J= Heat lost 11924J
Heat lost = 11924J = Mass x 4.184 x (82-96)
11924J / (4.184 x -14) = Mass
Mass = 203.56 = 204 mL (with proper significant figures)
Is this right?