Answer:
311.25k
Explanation:
The question assumes heat is not lost to the surroundings, therefore
heat emitted from hotter sample ( [tex]q_{\ lost}[/tex] )= heat absorbed by the less hotter sample( [tex]q_{\ gain}[/tex] )
The relationship between heat (q), mass (m) and temperature (t) is [tex]q = mc\Delta t[/tex]
where c is specific heat capacity, [tex]\Delta t[/tex] temperature change.
[tex]\Delta t[/tex] = [tex]t_{\ final} - t_{\ initial}[/tex]
equating both heat emitted and absorb
[tex]-q_{\ lost} = q_{\ gain}[/tex]
[tex]-m_{1}(t_{\ final} - t_{\ 1initial})=m_{2}(t_{\ final} - t_{\ 2initial})[/tex]
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
[tex]-m_{1}(t_{\ final} - t_{\ 1initial})=m_{2}(t_{\ final} - t_{\ 2initial})[/tex]
where m1 = 50g, t 1initial = 330, m2 = 30g, t2 initial = 280,t final (final temperature of the mixture) = ?
-50 * ([tex]t_{final}[/tex] - 330) = 30 * ([tex]t_{final}[/tex] - 280)
-50[tex]t_{final}[/tex] + 16500 = 30[tex]t_{final}[/tex] - 8400
80[tex]t_{final}[/tex] = 16500+8400
80[tex]t_{final}[/tex] = 24900
[tex]t_{final}[/tex] = 24900/80 = 311.25k
