In: Chemistry
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.
Given,
Volume of water(V1) = 330.0 mL
Initial temperature of water(Ti)1 = 25.00 oC
Volume of water(V2) = 100.0 mL
Initial temperature of water(Ti)2 = 95.00 oC
Final temperature of mixture(Tf) = ?
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 (Tf - 95.00) oC) = (330.0 g x (Tf - 25.00) oC)
-100.0 Tf + 9500 = 330.0 Tf - 8250
430 Tf = 17750
Tf = 41.28 oC
Thus, the final temperature of the mixture is 41.28 oC