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

insects!

explanation:

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s