Question

If you combine 250.0 mL of water at 25°C and 120.0 mL of water at 95°C, what is the final temperature of the mixture?

Answer 1

`T_f^oC = 47.07^oC`

Explanation:

We can use a heat equilibrium equation to find this value.

`DeltaH = m(T_f-T_i)C_p`

`DeltaH =` Change in Heat
`m =` Mass
`T=` Tempurature
`C_p=` Specific Heat

`DeltaH` cold `H_2O` = -`DeltaH` hot `H_2O`

`m(T_f-T_i)C_p = -[m(T_f-T_i)C_p]`

Since the density of water is `1 g/(mL)` the volume of water is also equal to the mass of the water in grams.

Cold `H_2O`

`m = 250.0g`
`T_f= ?`
`T_i = 25^oC`
`C_p = 4.18 J/(g^oC)`

Cold `H_2O`

`m = 120.0g`
`T_f= ?`
`T_i = 95^oC`
`C_p = 4.18 J/(g^oC)`

`m(T_f-T_i)C_p = -[m(T_f-T_i)C_p]`

`250.0g(T_f-25^oC)4.18 J/(g^oC) = -[120.0g(T_f-95^oC)4.18 J/(g^oC)]`

`1,045T_^oC - 26,125J/(g^oC) = -501.6T_f^oC + 47,652J/(g^oC)`

`(cancel(1,546.6)T_f^oC)/(cancel(1,546.6)) = (73,777)/(1,546.6)`

`T_f^oC = 47.07^oC`