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21 July, 06:44

A beaker of negligible heat capacity contains 456 g of ice at - 25.0°C. A lab technician begins to supply heat to the container at the rate of 1000 J/min. How long after starting will it take before the temperature starts to rise above 0°C

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  1. 21 July, 10:10
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    176 min

    Explanation:

    456 g =.456 kg

    Specific heat of ice s = 2093 J kg⁻¹

    Heat required to raise the temperature by 25 degree

    = mass x specific heat x rise in temperature.

    =.456 x 2093 x 25

    =23860 J

    Heat required to melt the ice to make water at zero degree

    = mass x latent heat

    =.456 x 334 x 10³

    =152304 J

    Total heat required = 152304 + 23860 = 176164 J.

    Time Required = Heat required / rate of supply of heat

    = 176164 / 1000

    176.16 min
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