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

Answer:

311.25k

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