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

t = 1/400 h   ⇔ 0.0025 h ⇔ 9s ⇔ 0.15 m

any of the answers are correct.