Question

A 30-0 g sample of water at 280 K is mixed with 50.0 g of water at 330 K. How would you calculate the final temperature of the mixture assuming no heat is lost to the surroundings?

Answer 1

311.25k

Step-by-step explanation:

The question assumes heat is not lost to the surroundings, therefore

heat emitted from hotter sample (
`q_(\ lost)` )= heat absorbed by the less hotter sample(
`q_(\ gain)` )

The relationship between heat (q), mass (m) and temperature (t) is
`q = mc\Delta t`

where c is specific heat capacity,
`\Delta t` temperature change.

`\Delta t` =
`t_(\ final) - t_(\ initial)`

equating both heat emitted and absorb

`-q_(\ lost) = q_(\ gain)`

`-m_(1)(t_(\ final) - t_(\ 1initial))=m_(2)(t_(\ final) - t_(\ 2initial))`

where the values with subset 1 are the values of the hotter sample of water and the values with subset 2 are the values of the less hot sample of water.

C will cancel out since both are water and they have the same specific heat capacity.

so we have

`-m_(1)(t_(\ final) - t_(\ 1initial))=m_(2)(t_(\ final) - t_(\ 2initial))`

where m1 = 50g, t 1initial = 330, m2 = 30g, t2 initial = 280,t final (final temperature of the mixture) = ?

-50 * (
`t_(final)` - 330) = 30 * (
`t_(final)` - 280)

-50
`t_(final)` + 16500 = 30
`t_(final)` - 8400

80
`t_(final)` = 16500+8400

80
`t_(final)` = 24900

`t_(final)` = 24900/80 = 311.25k