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

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