Question

When 131.0 mL of water at 26.0°C is mixed with 81.0 mL of water at 85.0°C, what is the final temperature? (Assume that no heat is lost to the surroundings; d of water is 1.00 g/mL.)

Answer 1

Answer:

63.52°C is the final temperature

Explanation:

1) 131.0 mL of water at 26.0°C

Mass of water = m

Volume of the water =131.0 mL

Density of the water = 1.00 g/mL

$Density=1.00 g/mL=\frac{m}{131.0 mL}$

m = 131.0 g

Initial temperature of the water = $T_i$ = 26.0°C

Final temperature of the water = $T_f$

Change in temperature ,$\Delta T=T_f-T_i$

Heat absorbed 131.0 g of water = Q

$Q=m\times c\times \Delta T$

2) 81.0 mL of water at 85.0°C

Mass of water = m'

Volume of the water =81.0 mL

Density of the water = 1.00 g/mL

$Density=1.00 g/mL=\frac{m'}{81.0 mL}$

m' = 81.0 g

Initial temperature of the water = $T_i'$ = 85.0°C

Final temperature of the water = $T_f'$

Change in temperature ,$\Delta T'=T_f'-T_i'$

Heat lost by 81.0 g of water = Q'

$Q'=m'\times c\times \Delta T'$

After mixing both liquids the final temperature will become equal fro both liquids.

$T_f=T_f'$

Since, heat lost by the water at higher temperature will be equal to heat absorbed by the water at lower temperature.

Q=-Q' (Law of conservation of energy.)

Let the specific heat of water be c

$m\times c\times \Delta T=m'\times c\times \Delta T'$

$131.0 g\times c(T_f-26^oC)=-(81.0.0 g\times c(T_f-85^oC))$

$T_f=63.52^oC$

63.52°C is the final temperature