Find the final equilibrium temperature when 15.0 g of milk at 13.0 degrees c is added to 148 g of coffee with a temperature of 88.3 degreesc. assume the specific heats of coffee and milk are the same as for water (c = 4.19 j/g•c), and disregard the heat capacity of the container.

Answers

Answer 1

Answer:

According to zeroth law of thermodynamics, when two objects are kept in contact, heat (energy) is transferred from one to the other until they reach the same temperature (are in thermal equilibrium). When the objects are at the same temperature there is no heat transfer.

So, at equilibrium, $q_(lost)$ = $q_(gain)$ , $q_(lost)= q_(milk)$ + $q_(coffee)$

q=m×c×T, where q = heat energy, m = mass of a substance, c = specific heat (units J/kg∙K), T is temperature

$q_(lost)= (15X13X4.19)+(148X88.3X4.19), (15+148)X4.19XT_(final)$ =(15X13X4.19)+(148X88.3X4.19)

$T_(final)$ = 81.37 ° C

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