Your 300 mL cup of coffee is too hot to drink when served at 87.0 degrees Celsius. What is the mass of the ice cube, taken from a - 15.0 degrees Celsius freezer, that will cool your coffee to a pleasant 65 degrees Celsius?

m=43.39 Kg

Explanation:

from conservation principle we have

heat gained by ice equal to heat lost by coffee

heat gained by ice = heat from - 15 °C ice to 0 °C ice + 0 °C ice to 0 °C water + 0 °C water to 65 °C water

= m*S1*15 + mL + ms2 * 65

where s is specific heat of respective phase

= m (2.08 * 15 + 334.16 + 4.184 * 65)

= m*637.32

heat lost by the coffee = ms * delta T

=0.3 * 4190 * (87-65)

=27654

therefore

27654 = m*637.32

m=43.39 Kg