limelight11
14.09.2019 •
Physics
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?
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Ответ:
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
Ответ:
d
Explanation:
I'm not really sure maybe