A sample of water with a mass of 648.00 kg at 298 k is heated with 87 kj of energy. the specific heat of water is 1 j-1 kg k-1.
what is the final temperature of the water?
1 answer:
The final temperature of the water is 298. 14 k
<h3>What is quantity of heat?</h3>
The formula for quantity of heat is given as:
Q = mc Δ T = m c ( final temperature - initial temperature)
Where:
- Q is Heat supply = 87 kj
- m is the mass of substance = 648.00 kg
- c is the specific heat capacity = 1 j-1 kg k-1
- ΔT is the temperature change
Substitute the values
87 = 648 × 1 ( T₂ - 298)
87 = 648 ( T₂ - 298)
Divide both sides by 648
87/ 648 = ( T₂ - 298)
0. 134 = T₂ - 298
Make 'T₂' subject of formula
T₂ = 298 + 0. 134
T ₂ = 298. 14 k
Thus, the final temperature of the water is 298. 14 k
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