Question

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?

Answer 1

The final temperature of the water is 298. 14 k

What is quantity of heat?

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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