Question

2.5kg of water (C = 4189 J/(Kg x K)) is heated from T1 = 13.5 C to T2 = 22.5 C

Answer 1

Answer:

338 K

Explanation:

$m_{1}$ = 2.5 kg, c = 4189 J/(kg K), $T_{1}$ = 13.5 $^{o} C$, $T_{2}$ = 22.5 $^{o} C$

$m_{2}$ = 0.5 kg, $T_{3}$ = 20 $^{o} C$

Heat loss by hotter water = heat gained by cooler water

$m_{1}$cΔT = $m_{2}$cΔT

$m_{1}$c($T_{2}$ - $T_{1}$) = $m_{2}$c($T_{4}$ - $T_{3}$)

2.5 x 4189 x (22.5 - 13.5) = 0.5 x 4189 x ($T_{4}$ - 20)

2.5 x 4189 x 9 = 2094.5 ($T_{4}$ - 20)

94252.5 = 2094.5$T_{4}$ - 41890

94252.5 + 41890 = 2094.5$T_{4}$

136142.5 = 2094.5$T_{4}$

$T_{4}$ = $\frac{136142.5}{2094.5}$

    = 65

$T_{4}$  = 65 $^{o} C$

But,

θ K = 273 + θ$^{o} C$

      = 273 + 65

      = 338

The final temperature of the water is 338 K.