# 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?

##### 1 Answer

#### Explanation:

If you assume that no heat is lost to the surroundings, then you can say that the heat **lost** by the *hotter* sample will be **equal to** the heat **absorbed** by the *cooler* sample.

Now, the equation that establishes a relationship between heat lost or gained and change in temperature looks like this

#color(blue)(q = m * c * DeltaT)" "# , where

*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 **closer** to

So, you can say that

#-q_"lost" = q_"gained"#

Here the minus signed is used because heat **lost** is **negative**.

This is equivalent to

#m_"hot" * color(red)(cancel(color(black)(c))) * DeltaT_"hot" = m_"cold" * color(red)(cancel(color(black)(c))) * DeltaT_"cold"#

Let's say that the final temperature is

#-50.0color(red)(cancel(color(black)("g"))) * (color(blue)(T_"f") - 330)color(red)(cancel(color(black)("K"))) = 30.0color(red)(cancel(color(black)("g"))) * (color(blue)(T_"f") - 280)color(red)(cancel(color(black)("K")))#

#-50.0 * color(blue)(T_"f") + 16500 = 30.0 * color(blue)(T_"f") - 8400#

#80.0 * color(blue)(T_"f") = 24900#

#color(blue)(T_"f") = 24900/80.0 = "311.25 K"#

Rounded to two sig figs, the number of sig figs you have for the two temperatures, the answer will be

#T_"f" = color(green)("310 K")#