Question

If 20 mL of water at 30°C is mixed with 60 mL of water at 80°C what will the final temperature of the water be?

Answer 1

67.5°C will be the final temperature of the water.

Step-by-step explanation:

Density of water = 1 g/ml

mass = Density × Volume

Mass of 20 mL water =
`m_1`

`m_1=1 g/ml* 20 mL=20 g`

Mass of 60 mL water =
`m_2`

`m_2=1 g/ml* 60 mL=60 g`

Heat gained by water at 30°C will be equal to heat lost by the water at 80°C

`Q_1=-Q_2`

Mass of water at 30°C=
`m_1=20 g`

Specific heat capacity of water =
`c_1=4.184 J/g^oC`

Initial temperature water at 30°C =
`T_1=30^oC`

Final temperature after mixing =
`T_2`=T

`Q_1=m_1c_1* (T-T_1)`

Mass of water at 80°C=
`m_2=60 g`

Specific heat capacity of water at 80°C=
`c_2=4.184 J/g^oC`

Initial temperature of the water at 80°C=
`T_3=80^oC`

Final temperature of water after mixing=
`T_2`=T

`Q_2=m_2c_2* (T-T_3)`

`Q_1=-Q_2`

`(m_1c_1* (T-T_1))=-m_2c_2* (T-T_3)`

`(m_1* (T-T_1))=-m_2* (T-T_3)`

On substituting all values:

`(20 g* (T-30^oC))=-[60 g* (T-80^oC)]`

we get, T = 67.5 °C

67.5°C will be the final temperature of the water.