Question

A 31.0-g sample of water at 290. k is mixed with 47.0 g water at 340. k. calculate the final temperature of the mixture assuming no heat loss to the surroundings.

Answer 1

We will use this formula:

M1C1ΔT1 = M2C2ΔT2

when M1 mass of sample 1 = 31 g

C1 for water = 4.184

and ΔT1 = (Tf- 290)

and M2 of sample 2 = 47 g

C2 for water = 4.184

and ΔT2 = (340 - Tf)

so by substitution, we can get Tf (the final temperature):

31 g * 4.184 * ( Tf - 290 k ) = 47g * 4.184 * (340 - Tf)

∴Tf ( the final temperature) = 320 K