If 50 ml of water at 85 degrees C are added to 100ml of water at 25 degrees C what would be the final temperature of the water assuming that no heat is lost to surroundings?

Answer 1

heat lost by warm water MUST equal heat gained by cooler water

q = mC∆T where q = heat; m = mass; C = specific heat and ∆T = change in temperature

For this problem, we will assume a density of water = 1 g / ml so we can convert volume to mass

C for water = 4.184 J/gº

For the warm water: q = mC∆T = (50 g)(4.184 J/gº)(85º - Tf) where Tf = final temperature

For the cooler water: q = mC∆T = (100 g)(4.184 J/gº)(Tf - 25º) where Tf = final temperature

Setting these two equal to each other, we have...

(50 g)(4.184 J/gº)(85º - Tf) = (100 g)(4.184 J/gº)(Tf - 25º) and since 4.184 is common to both sides....

(50 g)(85º - Tf) = (100 g)(Tf - 25º)

4250 - 50Tf = 100Tf - 2500

150Tf = 6750

Tf = 45ºC = final temperature