Question

A 90 ˚C liquid of mass 1 kg and specific heat capacity 4.5 x 103
J/kg K is poured into a
brass calorimeter of mass 150 g containing an unknown liquid of mass 20 g at a temperature
of 25 ˚C. If the final temperature of the mixture is 70 ˚C, determine the specific heat
capacity of the unknown liquid. The s.h.c of brass is 380 J/kg K.

Answer 1

$c_1 = 4500 \dfrac{J}{kg \cdot K}$ $m_1 = 1000g$ $T_1 = 90^\circ C$

$c_2 = 380 \dfrac{J}{kg \cdot K}$ $m_2 = 150g$ $T_2 = T_3 = 25^\circ C$

$c_3 = ?$ $m_3 = 20g$ $T = 70^\circ C$

$c_1 m_1(T - T_1) + c_2 m_2(T - T_2) + c_3 m_3(T - T_3) = 0$

$c_3 m_3(T - T_3) = c_1 m_1(T_1 - T) + c_2 m_2(T_2 - T)$

$c_3 = \dfrac{c_1 m_1(T_1 - T) + c_2 m_2(T_2 - T)}{m_3(T - T_3)} =$ $= \dfrac{ 4500 \dfrac{J}{kg \cdot K} \cdot 1000g \cdot (90^\circ C - 70^\circ C) + 380 \dfrac{J}{kg \cdot K} \cdot 150g \cdot (25^\circ C - 70^\circ C)}{20g \cdot (70^\circ C - 25^\circ C)} =$

$= 97150 \dfrac{J}{kg \cdot K}$

Please, check your task, because answer is a huge value for any substance. Then you should use my formulas with new values. Good luck!