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22 December, 23:07

What mass of water at 10.0 oC would be required to cool 50.0g of a metal having a specific heat of

0.60 J/g. oC from 90.0 oC to 20.0 oC?

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  1. 22 December, 23:12
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    The mass of water required is 50.2 g

    Explanation: Quantity of heat of a substance is calculated by multiplying the mass of a substance by the specific heat capacity and the change in temperature.

    That is;

    Q=m*c*ΔT

    In this case, water is used to cool a metal, therefore, water will gain heat while the metal will lose heat.

    Heat gained is equivalent to heat lost

    Heat gained by water = Heat lost by the metal

    Step 1: Heat lost by the metal

    Mass of the metal = 50.0 g

    Specific heat capacity of metal = 0.60 J/g°C

    Change in temperature (ΔT) = 70°C

    Heat lost by the metal = 50 g * 0.6 * 70

    = 2100 Joules

    Step 2: Heat gained by water

    Mass of water = x g

    Specific heat capacity of water = 4.18 J/g°C

    Temperature change = 10 °C

    Heat gained by water = x g * 4.18 * 10

    = 41.8x joules

    Step 3: Mass of water

    Heat gained by water = heat lost by the metal

    41.8x = 2100

    x = 50.239 g

    = 50.2 g (1 d. p)

    The mass of water is 50.2 g
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