Question

370 cm3 of water at 80°C is mixed with 120 cm3 of water at 4°C. Calculate the final equilibrium temperature, assuming no heat is lost to outside the water.

Answer 1

Answer : The final temperature of the mixture is
`61.4^oC`

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

`q_1=-q_2`

`m_1* c_1* (T_f-T_1)=-m_2* c_2* (T_f-T_2)`

And as we know that,

Mass = Density × Volume

Thus, the formula becomes,

`(\rho_1* V_1)* c_1* (T_f-T_1)=-(\rho_2* V_2)* c_2* (T_f-T_2)`

where,

`c_1` =
`c_2` = specific heat of water = same

`m_1` =
`m_2` = mass of water = same

`\rho_1` =
`\rho_2` = density of water = 1.0 g/mL

`V_1` = volume of water at
`80.0^oC` =
`370cm^3=370mL`

`V_2` = volume of water at
`4^oC` =
`120cm^3=120mL`

`T_f` = final temperature of mixture = ?

`T_1` = initial temperature of water =
`80.0^oC`

`T_2` = initial temperature of water =
`4^oC`

Now put all the given values in the above formula, we get:

`(\rho_1* V_1)* (T_f-T_1)=-(\rho_2* V_2)* (T_f-T_2)`

`(1.0g/mL* 370mL)* (T_f-80.0)^oC=-(1.0g/mL* 120mL)* (T_f-4)^oC`

`T_f=61.4^oC`

Therefore, the final temperature of the mixture is
`61.4^oC`