2.5kg of water (C = 4189 J/(Kg x K)) is heated from T1 = 13.5 C to T2 = 22.5 C To the heated water 0.5 kg of water at T3= 20 C is added. What is the final temperature of the water in Kelvin?

The value of the temperature in kelvin is 338 K.

The given parameters;

The final temperature of the water is calculated as follows as shown below;

The value of the temperature in kelvin is calculated as;

Learn more here:link

338 K

Explanation:

= 2.5 kg, c = 4189 J/(kg K), = 13.5 , = 22.5

= 0.5 kg, = 20

Heat loss by hotter water = heat gained by cooler water

cΔT = cΔT

c( - ) = c( - )

2.5 x 4189 x (22.5 - 13.5) = 0.5 x 4189 x ( - 20)

2.5 x 4189 x 9 = 2094.5 ( - 20)

94252.5 = 2094.5 - 41890

94252.5 + 41890 = 2094.5

136142.5 = 2094.5

=

    = 65

 = 65

But,

θ K = 273 + θ

      = 273 + 65

      = 338

The final temperature of the water is 338 K.

the answer to this question is 60 n