0.5000 kg of water at 35.00 degrees celsius is cooled, with the removal of 6.300 e4 j of heat. what is the final temperature of the water? specific heat capacity of water is 4186 j/(kg c°).

=30.22°C

Explanation:

The enthalpy change made the water to cool.

Enthalpy change = MC∅ where M is the mass of the water, C is the specific heat capacity of water and ∅ the temperature change.

ΔH=MC∅

∅=ΔH/MC

=(6.3×10⁴J)/(0.5kg×4186J/(kg°C))

=30.1

Final temperature = 35.00°C-30.1°C

=4.9°C

Explanation:

Given that,

Mass of water is m= 0.5kg

At initial temperature of θi = 35°C

Specific heat capacity of water

c = 4186J/kg•C

Heat H = 6.3 × 10⁴ J

Final temperature θf =?

Using the heat formulas

H = mc∆θ

H = mc(θf-θi)

Where,

H is heat

m is mass of substance

θi is initial temperature

θf is final temperature

Since it is cooled

H = mc(θi - θf)

6.3×10⁴ = 0.5×4186(35—θf)

63000/(0.5×4186) = (35-θf)

30.1 = 35-θf

-θf = 30.1-35

-θf = -4.9°C

θf = 4.9°C

The final temperature is 4.9°C

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