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6 September, 03:40

A cup holding 125.12g of water has an initial temperature of 26.8 degrees C. After a 35.08g piece of metal, at 99.5 degrees C, is transferred to the water, the temperature rises to 29.7 degrees C. What is the specific heat of the metal?

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  1. 6 September, 03:42
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    0.620 J/g°C

    Explanation:

    Heat gained or absorbed, Q by a substance is calculated by;

    Q = mass * specific heat capacity * Change in temperature

    In this case we are given;

    Mass of water = 125.12 g Initial temperature of water = 26.8 °C Initial temperature of the metal = 99.5°C Final temperature of the mixture = 29.7 °C

    We are required to calculate the specific heat capacity of the metal;

    Step 1 : Heat absorbed by water

    Specific heat capacity of water = 4.184 J/g°C

    Temperature change of water = 29.7 °C - 26.8°C

    = 2.9 °C

    But, Q = m*c*ΔT

    Thus, Heat = 125.12 g * 2.9°C * 4.184 J/g°C

    = 1518.156 Joules

    Step 2; Heat lost by the metal

    Specific heat capacity of the metal = x J/g°C

    Temperature change of the metal = 29.7 °C - 99.5°C

    = - 69.8 °C

    But, Q = mcΔT

    Therefore;

    Heat lost by the meatl = 35.08 g * x J/g°C * 69.8 °C

    = 2.448.584x Joules

    Step 3: C; aculating the specific heat capacity of the metal

    The heat gained by water is equal to the heat lost by the metal

    Therefore;

    1518.156 Joules = 2.448.584x Joules

    x = 1518.156 J : 2.448.584 J

    = 0.620 J/g°C

    Therefore, the specific heat of the metal is 0.620 J/g°C
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