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7 September, 03:29

An aluminium container of mass 100 GM contains 200gm of ice at - 20c heat is added to system at rate of 100 Cal/s find temp of system after 4minutes

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  1. 7 September, 04:05
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    1) First calculate the total heat added to the system:

    time 4 min * 60 s / min = 240 s

    Heat = rate * time = 100 cal/s * 240 s = 24,000 cal

    2) Calculate the change due to each element of the systemr:

    2.1 Aluminun container, Q Al = mass * Specific heat * ΔT =

    mass = 100 g

    Specific heat (from a table) = 0.22 cal / (g °C)

    ΔT = T - (-20°C)

    Q Al = 100 g * 0.22 cal / (g°C) * (T + 20) = 22 (T + 20)

    2.2 Q ice from - 20 °C to 0°C

    mass = 200 g

    Specific heat = 0.49 cal / (g°C)

    ΔT = (0°C - (-20°C)) = 20°C

    2.3 Q ice = 200g * 0.49 cal / (g°C) * 20°C = 1,960 cal

    2.4 Latent fussion heat of ice

    From a table: 80 cal / g

    Q Lf = 200 g * 80 cal / (g) = 16,000 cal

    3) Heating of the water from 0°C to final T

    Specific heat = 1 cal / (g°C)

    Q w = 200g * 1 cal / (g°C) * (T - 0) = 200T

    5) Final calculus

    24,000 cal = Q Al + Q ice + Q Lf + Qw =

    = 22 (T + 20) + 1960 cal + 16,000 cal + 200T = 22T + 440 + 1960 + 200T =

    24,000 = 222T + 2400 = > 222T = 24000 - 2400 = 21,600 = >

    T = 21,600 / 222 = 97.3 °C

    Answer: 97.3 °C
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