A 30-0 g sample of water at 280 K is mixed with 50.0 g of water at 330 K. How would you calculate the final temperature of the mixture assuming no heat is lost to the surroundings?
Answer:
\(310 \mathrm{~K}\)
Explanation:
If you assume that no heat is lost to the surroundings, then you can say that the heat \(\text{lost}\) by the hotter sample will be \(\text{equal to}\) the heat \(\text{absorbed}\) by the cooler sample.
Now, the equation that establishes a relationship between heat lost or gained and change in temperature looks like this
\(q=m \cdot c \cdot \Delta T\), where
\(q\) – heat lost or gained
\(m\) – the mass of the sample
\(c\) – the specific heat of the substance
\(\Delta T\) – the change in temperature, defined as final temperature minus initial temperature
Now, before doing any calculation, try to predict what will happen when the two samples are mixed.
Notice that you have more hot water than cold water, which means that you can expect the final temperature of the mixture to be \(\text{closer}\) to \(330 K\) than to \(280 K\).
So, you can say that
\(-q_{\text {lost }}=q_{\text {gained }}\)
Here the minus signed is used because heat \(\text{lost}\) is \(\text{negative.}\)
This is equivalent to
\(m_{\text {hot }} \cdot \not c \cdot \Delta \cdot \Delta T_{\text {hot }}=m_{\text {cold }} \cdot \not c \cdot \Delta T_{\text {cold }}\)
\(-50.0 \mathrm{\not g} \cdot\left(T_{\mathrm{f}}-330\right) \mathrm{\not K}=30.0 \mathrm{\not g} \cdot\left(T_{\mathrm{f}}-280\right) \mathrm{\not K}\)
\(-50.0 \cdot T_{\mathrm{f}}+16500=30.0 \cdot T_{\mathrm{f}}-8400\)
\(80.0 \cdot T_{\mathrm{f}}=24900\)
\(T_{\mathrm{f}}=\frac{24900}{80.0}=311.25 \mathrm{~K}\)
Rounded to two sig figs, the number of sig figs you have for the two temperatures, the answer will be
\(T_{\mathrm{f}}=310 \mathrm{~K}\)Free, Modern, efficient, accurate and user friendly Specific Heat Calculator.