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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