Question

Show the calculation of the final temperature of the mixture when a 40.5 gram sample of water at 85.7 degrees celcius is added to 36.8 gram sample of water at 26.3 degree celcius in a coffee cup calorimeter

Answer 1

The value of final temperature of the mixture
`T_(3)` = 57.42 °c

Step-by-step explanation:

Mass of first sample
`m_(1)` = 40.5 gram

Mass of second sample
`m_(2)` = 36.8 gram

Temperature of first sample
`T_(1)` = 85.7 °c

Temperature of second sample
`T_(2)` = 26.3 °c

The energy balance equation for the mixture is written as

`m_(1)`
`T_(1)` +
`m_(2)`
`T_(2)` =
`m_(3)`
`T_(3)` ---------- ( 1 )

Total mass
`m_(3)` =
`m_(1)` +
`m_(2)`

Put the values of
`m_(1)` &
`m_(2)` in this equation we get,

⇒
`m_(3)` = 40.5 + 36.8

⇒
`m_(3)` = 77.3 gram -------- (2)

Put all the values of mass & temperature in equation 1 we get,

⇒ 40.5 × 85.7 + 36.8 × 26.3 = 77.3 ×
`T_(3)`

⇒ 4438.69 = 77.3 ×
`T_(3)`

⇒
`T_(3)` = 57.42 °c

This is the value of final temperature of the mixture.