Question

Question 4 of 15 > If you combine 260.0 mL of water at 25.00 °C and 130.0 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. Tal

Answer 1

sol:-

data provided in the question is :-

• Volume of first water = 260 mL
• temperature of first water = 25 oC
• voulume of second water = 130.0 mL
• temperature of second water = 95 oC
• density of water = 1.00 g/mL

mass of first water (m1) = (density of water) * ( Volume of water)  

=( 1 g/mL ) * (260 mL) = 260 g

mass of second water(m2) = (1 g/mL) * (130 mL)

= 130 g

$\Delta$ H = mCp(dT)

let the final temperature = Tf oC

so,

heat lost by second water = heat gained by first water

- m2Cp(Tf - 95 oC) = m1Cp(Tf - 25 oC)

- m2(Tf - 95 oC) = m1(Tf - 25 oC)

- 130 g ( Tf - 95 oC) = 260(Tf - 25 oC)

- (Tf - 95 oC) = 2(Tf - 25 oC)

- Tf + 95 oC = 2Tf - 50 oC

2Tf + Tf = (95 + 50) oC

3Tf = 145 oC

Tf = (145/3) oC

= 48.33 oC

Tf = 48.33 oC