How many kilojoules of heat are absorbed when 1.00 L of water is heated from 18°C to 85°C? (Specific heat capacity of water is 4
.18 J/g°C, the density of water is 1 g/mL)
1 answer:
ANSWER: 280.06 kilojoules.
EXPLANATION:
Heat absorbed can be calculated from the equation Q = m . C . ΔT
where m = mass of substance in grams
C = specific heat capacity of substance in J/g°C
ΔT = temperature difference.
So, to calculate heat absorbed by water we must know three terms, mass of water, specific heat capacity of water and temperature difference.
Have a look at question, Specific heat capacity of water, C = 4.18 J/g°C.. Temperature difference is, ΔT = (85 - 18) °C = 67 °C. Mass of water is not given, instead we are given volume of water = 1 L and density of water = 1 g/mL. So we need to calculate mass of water from volume and density.
We know the formula, density is mass upon volume. see pic for calculation. mass of water comes out to be 1000 g.
So, now we know m = 1000 g, C = 4.18 J/g°C and ΔT = 67 °C.
Use, Q = m . C . ΔT to calculate heat absorbed.
Q = 1000 g . 4.18 J/g°C . 67 °C = 280060 J.
The answer comes out be in units of Joules but we need answer in kilojoules so divide the answer by 1000(thousand).
answer will be 280.06 kilojoules.
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