Question

Assignment Score: 495/2800 Resources x Give Up? Hint Check Answer Question 13 of 28 > If you combine 330.0 mL of water at 25.00 °C and 110.0 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. Tfinal =

Answer 1

volume of hot water =110ml

volume of cold water =330ml

density =mass/volume

temperature of copper Thot water =95⁰C

temperature of water T_{cold water} =25⁰C

let the equilibrium temperature be T

density =1g/ml

mass of hot water m_hotwater =density * volume

=1g/ml*110ml

=110g

mass of cold water m_coldwater =density * volume

=1g/ml*330ml

=330g

specific capacity of water C_water=4.183 J^oC^-1g

heat gained by cold body $q^{water}=m_{water}C_{water}\Delta T=m_{water}C_{water}(T-T_{water})$

q_coldater = 330 g *4.183 JC^-1g*(T-25)

heat lost by hot body $q^{water}=m_{water}C_{water}\Delta T$

q_hotwater= 110 g *4.183 J^oC^-1g(95-T)

by principle of calorimetry

heat gained by cold body =heat lost by hot body

q_coldwater =q_hotwater

330 g *4.183 JC^-1g*(T-25)=110 g *4.183 J^oC^-1g(95-T)

cancelling 4.183 from both side

330 *(T-25)=110* (95-T)

dividing both side by 110

3 *(T-25)= (95-T)

3T-75=95-T

3T+T=95+75

4T=170

T=170/4 =42.5^0C

T=42.5⁰C