A 33.05 g sample of a substance is initially at 28.5 °C. After absorbing 2589 J of heat, the temperature of the substance is 107.9 ‘C.

Question

A 33.05 g sample of a substance is initially at 28.5 °C. After absorbing 2589 J of heat, the temperature of the substance is 107.9 ‘C.
What is the specific heat (c) of the substance?

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Dâu 1 month 2021-08-15T04:39:09+00:00 1 Answers 1 views 0

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    2021-08-15T04:40:51+00:00

    Answer:

    0.985\ \text{J/g}^{\circ}\text{C}

    Explanation:

    m = Mass of sample = 33.05 g

    \Delta T = Change in temperature = 107.9-28.5=79.5^{\circ}\text{C}

    Q = Heat absorbed = 2589 J

    c = Specific heat of substance

    Heat is given by

    Q=mc\Delta T\\\Rightarrow c=\dfrac{Q}{m\Delta T}\\\Rightarrow c=\dfrac{2589}{33.05\times 79.5}\\\Rightarrow c=0.985\ \text{J/g}^{\circ}\text{C}

    The specific heat of the substance is 0.985\ \text{J/g}^{\circ}\text{C}.

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