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17 September, 16:31

If 20 kg of iron, initially at 12 °C, is added to 30 kg of water, initially at 90 °C, what would be the final temperature of the combined system? (Hint: the heat given up by the water will be equal to the heat gained by the iron) Explain how you would represent this problem in the simulation.

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  1. 17 September, 16:41
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    final temperature of the combined system T = 84.78°C

    Explanation:

    Given data

    mass of iron (m1) = 20 kg

    temperature iron (t1) = 12 °C

    mass of water (m2) = 30 kg

    temperature of water (t2) = 90 °C

    To find out

    final temperature of the combined system

    solution

    we know the energy requirement formula to rise the temp

    energy = mass * specific heat * change in temperature

    we combine both system so both energy will be added

    and

    we know specific heat of iron (c1) = 0.450 kJ/kg

    and specific heat of water (c2) = 4.186 kJ/kg

    4.186 joule/gram °C

    now combine both energy

    energy = mass, m1 * specific heat, c1 * change in temperature, T - t1 + mass, 2 * specific heat, c2 * change in temperature, T - t2

    energy = 20 * 0.450 * T - 12 + 30 * 4.186 * T - 90

    (20) (0.45) (T-12) = (30) (4.186) (90-T)

    final temperature of the combined system T = 84.78°C
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