The specific heat of water is 4.184Jg ∘C. Determine the final temperature when 600.0 g water at 75.5∘C absorbs 5.90×104 J of energy.]

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The specific heat of water is 4.184Jg ∘C. Determine the final temperature when 600.0 g water at 75.5∘C absorbs 5.90×104 J of energy.]

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Thanh Hà 3 years 2021-08-29T12:57:56+00:00 1 Answers 1 views 0

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  1. Cherry
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    2021-08-29T12:59:15+00:00
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    Answer:

    T_2=98.5^{\circ}

    Explanation:

    Given that,

    The specific heat of water is 4.184Jg°C

    Mass, m = 600 g

    Initial temperature, T₁ = 75.5°C

    We need to find the final temperature. We know that heat absorbed is given by :

    Q=mc\Delta T\\\\Q=mc(T_2-T_1)\\\\\dfrac{Q}{mc}=(T_2-T_1)\\\\\\T_2=\dfrac{Q}{mc}+T_1\\\\T_2=\dfrac{5.9\times 10^4}{600\times 4.184}+75\\\\T_2=98.5^{\circ}

    So, the final temperature is equal to 98.5^{\circ}.

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