Question #235836
how much energy must be absorbed by 20.0g of water to increase tempature from 283.0
to 303.0
Expert's answer
Q = C x m deltaT
deltaT = 303 - 283 = 20
C = 4.2 Joules per gram per degree Celsius (J/g°C)
Q = 4.2 x 20 x 20 = 1 680 J
how much energy must be absorbed by 20.0g of water to increase tempature from 283.0
to 303.0
Q = C x m deltaT
deltaT = 303 - 283 = 20
C = 4.2 Joules per gram per degree Celsius (J/g°C)
Q = 4.2 x 20 x 20 = 1 680 J
