What is the mass of water?
Express your answer using two significant figures.
Use the equation
Q = mcDeltaT
Q = 1.3 x 10^6 J
m = to be found
c = specific heat of water 4.12
DeltaT - temperature difference ( 100 - 13 = 87)
Substituting
1.3 x 10^6 = m X 4.12 x 87
m(g) = 1.3 x 10^6 / (4.12 x 87)
m(g) = 3626.82 g = 3.626 kg ( volume would be 3.626 dm^3 (litre).
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Verified answer
Use the equation
Q = mcDeltaT
Q = 1.3 x 10^6 J
m = to be found
c = specific heat of water 4.12
DeltaT - temperature difference ( 100 - 13 = 87)
Substituting
1.3 x 10^6 = m X 4.12 x 87
m(g) = 1.3 x 10^6 / (4.12 x 87)
m(g) = 3626.82 g = 3.626 kg ( volume would be 3.626 dm^3 (litre).