Respuesta :
Given:
The initial temperatures of the liquids 1,2 and 3 are
[tex]\begin{gathered} T_1=11\text{ C} \\ T_2=10\text{ C} \\ T_3=29\text{ C} \end{gathered}[/tex]the final temperature of the first two liquids is
[tex]T_{f1}=14\text{ C}[/tex]the final temperature of the second and third liquids is
[tex]T_{f2}=24.5\text{ C}[/tex]Required:
final temperature of the first and third liquidd.
Explanation:
to solve this problem apply principal of calorimetry here,
for the first two liquids is
[tex]\begin{gathered} \Delta Q=0 \\ mk_1(T_{f1}-T_1)+mk_2(T_{f1}-T)=0 \end{gathered}[/tex]plugging all the values in the above relation we get
[tex]\begin{gathered} k_1(14-11)=-k_2(14-19) \\ k_1=\frac{5}{3}k_2..........(1) \end{gathered}[/tex]here k1 nad k2 are the specific heat of the two liquids
now similarly for, second and third liquids
[tex]\begin{gathered} k_2(24.5-19)=-k_3(24.5-29) \\ k_2=\frac{4.5}{5.5}k_3 \\ k_2=\frac{9}{11}k_3\text{ ............}(2) \end{gathered}[/tex]here, k3 is specific heat of liquid three
from the equation (1) and (2) we can write
[tex]\begin{gathered} k_1=\frac{5}{3}\times\frac{9}{11}k_3 \\ k_1=\frac{15}{11}k_3 \\ \frac{k_{\frac{1}{}}}{k_3}=\frac{15}{11} \end{gathered}[/tex]now for liquid first and third
[tex]\begin{gathered} \Delta Q=0 \\ mk_1(T-T_1)+mk_3(T-T_3) \\ k_1T+k_3T=k_1T_1+k_3T_3 \\ T=\frac{\frac{k_1}{k_3}T_1+T_3}{\frac{k_1}{k_3}+1} \end{gathered}[/tex]Plugging all the values in the above relation we get
[tex]\begin{gathered} T=\frac{\frac{15}{11}\times11+29}{\frac{15}{11}+1} \\ T=\frac{44}{26}\times11 \\ T=18.61\text{ C} \end{gathered}[/tex]Thus, the final equilibrium temperature is
[tex]18.6\text{1 C}[/tex]