Question

A 49.0 g sample of water at 100. °C is poured into a 55.0 g sample of water at 25 °C. What will be the final temperature of the water? The specific heat of water is 4.184 J/g °C.

Final temperature =

Answer 1

To solve this problem, we can use the equation:

m1c1ΔT1 + m2c2ΔT2 = 0

where m1 and m2 are the masses of the two samples of water, c is the specific heat of water, and ΔT is the change in temperature. We can assume that the final temperature of the water is T.

Plugging in the values, we get:

(49.0 g)(4.184 J/g °C)(T - 100. °C) + (55.0 g)(4.184 J/g °C)(T - 25. °C) = 0

Simplifying and solving for T, we get:

T = [ (49.0 g)(4.184 J/g °C)(100. °C) + (55.0 g)(4.184 J/g °C)(25. °C) ] / [ (49.0 g)(4.184 J/g °C) + (55.0 g)(4.184 J/g °C) ]

T = 39.4 °C

Therefore, the final temperature of the water is 39.4 °C.