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21 April, 17:59

An immersion heater of power J = 500 W is used to heat water in a bowl. After 2 minutes, the temperature increases from T1 = 85°C to T2 = 90°C. The heater is then switched off for an additional minute, and the temperature drops by Estimate the mass of the water in the bowl. The thermal capacity of water c = 4.2 x 103 J K-1 Kg-1

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  1. 21 April, 20:40
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    Answer: 0.051kg

    Explanation:

    Using the law of calorimetry which states that the amount of heat lost by a hot substance is equal to the amount of heat gained by the cold substance.

    Applying the law to this question,

    Amount of heat lost by the heater = heat gained by the water

    Heat lost by Immersion heater Q1 = Power*time

    Given power = 500W time = 2minutes = 120seconds

    Heat loss by immersion heater = 500*120 = 60,000Joules

    Heat gained by water = mc∆t where;

    m is the mass of the water in the bowl = ?

    c is the specific heat capacity = 4.2x10³JK-1 Kg-1

    ∆t = change in temperature = 90°C - 85°C = 5°C

    = 273+5 = 278K

    Substituting in the formula we have;

    Q2 = 4.2*10³*278*m = 1,167,600m

    According to the law if calorimetry,

    Q2=Q1

    1,167,600m = 60000

    m = 60000/1,167,600

    m = 0.051kg

    Therefore the mass of water in the bowl is 0.051kg
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