Question

Three liquids are at temperatures of 11 degrees Celsius, 19 degrees Celsius, and 29 degrees Celsius respectively. Equal masses of the first two liquids are mixed, and the equilibrium temperature is 14 degrees Celsius. Equal masses of the second and third are then mixed, and the equilibrium temperature is 24.5 degrees Celsius. Find the equilibrium temperature when equal masses of the first and third are mixed. Answer in units of degrees Celsius.

Answer 1

Given:

The initial temperatures of the liquids 1,2 and 3 are

`\begin{gathered} T_1=11\text{ C} \\ T_2=10\text{ C} \\ T_3=29\text{ C} \end{gathered}`

the final temperature of the first two liquids is

`T_(f1)=14\text{ C}`

the final temperature of the second and third liquids is

`T_(f2)=24.5\text{ C}`

Required:

final temperature of the first and third liquidd.

Step-by-step explanation:

to solve this problem apply principal of calorimetry here,

for the first two liquids is

`\begin{gathered} \Delta Q=0 \\ mk_1(T_(f1)-T_1)+mk_2(T_(f1)-T)=0 \end{gathered}`

plugging all the values in the above relation we get

`\begin{gathered} k_1(14-11)=-k_2(14-19) \\ k_1=(5)/(3)k_2..........(1) \end{gathered}`

here k1 nad k2 are the specific heat of the two liquids

now similarly for, second and third liquids

`\begin{gathered} k_2(24.5-19)=-k_3(24.5-29) \\ k_2=(4.5)/(5.5)k_3 \\ k_2=(9)/(11)k_3\text{ ............}(2) \end{gathered}`

here, k3 is specific heat of liquid three

from the equation (1) and (2) we can write

`\begin{gathered} k_1=(5)/(3)*(9)/(11)k_3 \\ k_1=(15)/(11)k_3 \\ \frac{k_{\frac{1}{}}}{k_3}=(15)/(11) \end{gathered}`

now for liquid first and third

`\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=((k_1)/(k_3)T_1+T_3)/((k_1)/(k_3)+1) \end{gathered}`

Plugging all the values in the above relation we get

`\begin{gathered} T=((15)/(11)*11+29)/((15)/(11)+1) \\ T=(44)/(26)*11 \\ T=18.61\text{ C} \end{gathered}`

Thus, the final equilibrium temperature is

`18.6\text{1 C}`