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2 years ago

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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:

9966 [12]2 years ago

4 0

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

Learn more about quantity of heat here:

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<h3><u>Let's</u><u> </u><u>understand</u><u> </u><u>the concept</u><u>:</u><u>-</u></h3>

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