Answer to Question #167326 in Physical Chemistry for NOOR

Question

Determine the final temperature when 100 g of water at 20.0 °C mixes with 400 g of water at 80 °C and with 500 g of water at 60.0 °C 

Answer 1

M₁= 100g

T₁=20°C

M₂= 400g

T₂= 80°C

M₃= 500g

T₃= 60°C

T= ?

Heat gained = Heat lost

M₁(T-T₁) = M₂(T₂-T) + M₃(T₃-T)

100(T-20) = 400(80-T) + 500(60-T)

100T + 400T + 500T = 32000 + 30000 + 2000

1000T = 64000

T= 64°C