What mass of water can be heated from 25 degrees C to 75.0 degrees C by the addition of 7.96 x 10 4 Joules?

Question

What mass of water can be heated from 25 degrees C to 75.0 degrees C by the addition of 7.96 x 10 4 Joules?

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Kennedy 5 months 2021-07-18T23:50:48+00:00 2 Answers 0 views 0

Answers ( )

    0
    2021-07-18T23:52:06+00:00

    Answer:

    27.1 g is the correct answer

    0
    2021-07-18T23:52:45+00:00

    Explanation:

    Specific heat of water is = 1cal/°C = 4.18J/K

    That means water requires 4.18 Joules of heat energy to raise the temperature of 1 g of water by 1°C or 1K.

    Similarly water will lose 4.18 Joules of heat energy for every gram of water by 1°C.

    Therefore (625 x 4.18) Joules of heat energy is lost for every drop in 1°C of water.

    So number degrees it gets cooled for 7.96 x 10⁴ Joules

    = 79600/(625 x 4.18)

    =30.47°C

    So the final temperature of 625 g of water = (75–30.47)= 44.53°C

    c=4200 m=0.625 t1=75 dQ=7,96*10^4=79600J

    dt=t1-t2 d-delta(change) Q=cm(dt)

    dt=Q/cm=79600/(4200*0.625)=30.3238

    t2=75-30,3=44,7°C

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