When 565g of a metal with temperature of 90.00 C added to 200.0 g of water at 25.00 degrees. Assuming all the heat lost by the iron is trans

Question

When 565g of a metal with temperature of 90.00 C added to 200.0 g of water at 25.00 degrees. Assuming all the heat lost by the iron is transferred to the water ( specific heat capacity of water = 4/18 J/(g-C)), and the final temperature of the mixture is 34.40 degree Celsius what is the heat capacity of the metal?

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Euphemia 4 years 2021-08-30T15:36:01+00:00 1 Answers 8 views 0

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  1. Acacia
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    2021-08-30T15:37:19+00:00
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    Answer:

    The specific heat is 0.25J/g°C

    Explanation:

    To solve this question we must know that the heat released for the metal is equal to the heat absorbed for the water. The equation is:

    S(Metal)*m(Metal)*ΔT(metal) = S(water)*m(water)*ΔT(water)

    Where S of metal is our incognite (Specific heat of metal)

    m is the mass of the metal = 565g

    ΔT is change in heat of the metal = 90.00°C – 34.40°C = 55.60°C

    Specific heat of water = 4.18J/g°C

    The mass of water = 200.0g

    ΔT of water is 34.40°C – 25.00°C = 9.40°C

    Replacing:

    S(Metal)*565g*55.60°C = 4.18J/g°C*200.0g*9.40°C

    S(metal) = 0.25J/g°C

    The specific heat is 0.25J/g°C

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