Question

If you combine 330.0 mL of water at 25.00 °C and 100.0 mL of water at 95.00 °C, what is the final temperature of the mixture? Use 1.00 g/mL as the density of water.

Answer 1

Given,

Volume of water(V₁) = 330.0 mL

Initial temperature of water(T_i)₁ = 25.00 ^oC

Volume of water(V₂) = 100.0 mL

Initial temperature of water(T_i)₂ = 95.00 ^oC

Final temperature of mixture(T_f) = ?

Density of water(D) = 1.00 g/mL

Firstly converting the volume of water to mass,

= 330.0 mL x ( 1.00 g/ 1.00 mL)

m₁ = 330.0 g

Similarly,

m₂ = 100.0 mL x ( 1.00 g/ 1.00 mL)

m₂ = 100.0 g

Now, after combining 330.0 g water at 25.00 ^oC and 100.0 g of 95.00 ^oC,

Heat lossed by 100.0 g of water = Heat gained by 330.0 g of water

- (m₂ x C x T) = (m₁ x C x T)

Here, specific heat capacity of water = 4.18 J/ g ^oC

As "C" value is same for both sides,

- (m₂ x T) = (m₁ x T)

- (100.0 g x (T_f - 95.00) ^oC) = (330.0 g x (T_f - 25.00) ^o​C)

-100.0 T_f + 9500 = 330.0 T_f - 8250

430 T_f = 17750

T_f = 41.28 ^oC

Thus, the final temperature of the mixture is 41.28 ^oC