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24 June, 05:52

An electric range burner weighing 683.0 grams is turned off after reaching a temperature of 477.6°C, and is allowed to cool down to 23.2°C.

Calculate the specific heat of the burner if all the heat evolved from the burner is used to heat 552.0 grams of water from 23.2°C to 80.3°C.

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  1. 24 June, 05:56
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    0.102 cal/g.°C

    Explanation:

    According to the law of conservation of energy, the sum of the heat released by the electric burner (Qb) and the heat absorbed by the water (Qw) is zero.

    Qb + Qw = 0

    Qb = - Qw

    Both heats can be calculated using the following expression.

    Q = c * m * ΔT

    where,

    c: specific heat

    m: mass

    ΔT: change in the temperature

    Then,

    Qb = - Qw

    cb * mb * ΔTb = - cw * mw * ΔTw

    cb = - cw * mw * ΔTw / mb * ΔTb

    cb = - (1 cal/g.°C) * 552.0 g * (80.3°C - 23.2°C) / 683.0 g * (23.2°C - 477.6°C)

    cb = 0.102 cal/g.°C
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