An 80g ice cube at -10°C is placed in 680g of water at 24°C. What is the final temperature of this mixture?

Answer 1

To solve this problem, we can use the principle of conservation of energy:

m1c1ΔT1 + m2c2ΔT2 = 0

where
m1 = mass of ice cube = 80g
c1 = specific heat capacity of ice = 2.09 J/g°C
ΔT1 = final temperature of ice = T - (-10) = T + 10

m2 = mass of water = 680g
c2 = specific heat capacity of water = 4.18 J/g°C
ΔT2 = final temperature of water = T - 24

Setting up the equation with the known values:

80g * 2.09 J/g°C * (T + 10) + 680g * 4.18 J/g°C * (T - 24) = 0

Solving for T:

160.8(T + 10) + 2849.6(T - 24) = 0
160.8T + 1608 + 2849.6T - 68390.4 = 0
3010.4T - 66782.4 = 0
3010.4T = 66782.4
T ≈ 22.2°C

Therefore, the final temperature of the mixture is approximately 22.2°C.