Abeaker 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

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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snell's law

refractve index = sin incidence / sin refraction