# Answer to Question #5463 in Molecular Physics | Thermodynamics for Mackenzie

Question #5463

If 6.50 kJ are used to heat 63.0 grams of water at 20 degrees Celsius, what will be its final temperature?

Expert's answer

m = 63 g = 0,063 kg

Q = 6,5 kJ = 6500 J

с = 4200 J*kg-1*K-1

t1 = 20°C = 293 K

t2 - ?

Q = c*m*(t2-t1)

6500 = 4200*0,063*(t2-293)

24,565 = t2-293

t2 = 317,565 K

t2=317,6 K = 44,6°C

Q = 6,5 kJ = 6500 J

с = 4200 J*kg-1*K-1

t1 = 20°C = 293 K

t2 - ?

Q = c*m*(t2-t1)

6500 = 4200*0,063*(t2-293)

24,565 = t2-293

t2 = 317,565 K

t2=317,6 K = 44,6°C

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