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2 months ago

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a sample of water with a mass of 648.00 kg at 298 K is heated with 87 kh of energy. the specific heat of water is 1 J-1 kg K-1.

what is the final temperature of the water

2 answers:

lions [2.9K]2 months ago

8 0

The final temperature of the water is 432.26 K

Working out the solution:

Formula employed:

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

Where,

Q = heat applied = 87 kJ = 87000 J

m = water mass = 648.00 kg

c = water's specific heat = $1J/kg.K$

$\Delta T=\text{Change in temperature}$

$T_{final}$ = final temperature =?

$T_{initial}$ = initial temperature = 298 K

By substituting the known values into the formula, we can determine the final temperature of the water.

$87000J=648.00kg\times 1J/kg.K\times (T_{final}-298K)$

$T_{final}=432.26K$

Thus, the final temperature of the water is 432.26 K

Anarel [2.9K]2 months ago

3 0

q = mCΔT

The specific heat capacity for water is 4.187 kJ/(kg.K).

ΔT can be found using q/mC = 87 kJ/[648.00 kg x 4.187 kJ/(kg.K)] = 87 kJ/(2713 kJ/K) = 0.032 K

Final temperature Tf = Ti + ΔT = 298 K + 0.032 K = 298.032 K

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