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two portions of the same liquid are mixed together. one has a mass of 100.0 g and a temperature of 57 °c. the other has a mass of 25 g and a temperature of 21 °c. ignoring the container in which they are mixed and any heat lost from that container, determine the final temperature $t_f$ at equilibrium.
$t_f$ = number
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Explanation:

Step1: Apply the principle of heat transfer

The heat lost by the warmer liquid is equal to the heat gained by the cooler liquid. The formula for heat transfer is \(Q = mc\Delta T\), where \(m\) is mass, \(c\) is specific heat capacity (since it's the same liquid, \(c\) cancels out), and \(\Delta T\) is the change in temperature.
Let \(T_f\) be the final temperature.
For the warmer liquid: \(Q_1=m_1c(T_1 - T_f)\)
For the cooler liquid: \(Q_2=m_2c(T_f - T_2)\)
Since \(Q_1 = Q_2\), we have \(m_1(T_1 - T_f)=m_2(T_f - T_2)\)

Step2: Substitute the given values

Given \(m_1 = 100.0\space g\), \(T_1 = 57^{\circ}C\), \(m_2 = 25\space g\), \(T_2 = 21^{\circ}C\)
\(100(57 - T_f)=25(T_f - 21)\)
Expand: \(5700-100T_f = 25T_f-525\)

Step3: Solve for \(T_f\)

Add \(100T_f\) to both sides: \(5700=125T_f - 525\)
Add \(525\) to both sides: \(5700 + 525=125T_f\)
\(6225=125T_f\)
\(T_f=\frac{6225}{125}=49.8^{\circ}C\)

Answer:

\(49.8^{\circ}C\)