# Question #da1ce

##### 1 Answer

#### Explanation:

The idea here is that the heat **lost** by the hotter liquid will be **equal to** the heat **gained** by the colder liquid.

#color(blue)(q_"gained" = -q_"lost")" " " "color(purple)((1))#

Here the minus sign is used because *heat lost* carries a negative sign.

Since tea has the higher specific heat value and the higher initial temperature, you can expect the final temperature of the mixture to be **much closer** to the initial temperature of the tea.

As you know, a substance's specific heat tells you how much heat must be added or removed from a

In this case, tea has a higher specific heat than milk, which means that it needs to lose or gain **more heat** to produce a

The equation you're going to use looks like this

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

*final temperature* minus *initial temperature*

Now, you're going to have to convert the two specific heats from *Joules per kilogram Kelvin*, which is **equivalent** to *Joules per kilogram Celsius*, to *Joules per gram Celsius*.

To do that, use the conversion factor

#"1 kg" = 10^3"g"#

This will get you

#c_"milk" = 3800"J"/(color(red)(cancel(color(black)("kg"))) ""^@"C") * (1color(red)(cancel(color(black)("kg"))))/(10^3"g") = 3.800"J"/("g" ""^@"C")#

and

#c_"tea" = 4200"J"/(color(red)(cancel(color(black)("kg"))) ""^@"C") * (1color(red)(cancel(color(black)("kg"))))/(10^3"g") = 4.200"J"/("g" ""^@"C")#

Let's say that the final temperature of the mixture will be **change in temperature** for the substances will be

#DeltaT_"milk" = color(blue)(T_"f") - 20^@"C"" "# and#" " DeltaT_"tea" = color(blue)(T_"f") - 80^@"C"#

Plug your values into equation

#overbrace(m_"milk" * c_"milk" * DeltaT_"milk")^(color(red)(q_"milk")) = - overbrace(m_"tea" * c_"tea" * DeltaT_"tea")^(color(red)(q_"tea"))#

In your case, this will be equal to

#50 color(red)(cancel(color(black)("g"))) * 3.800color(red)(cancel(color(black)("J")))/(color(red)(cancel(color(black)("g"))) color(red)(cancel(color(black)(""^@"C")))) * (color(blue)(T_"f") - 20)color(red)(cancel(color(black)(""^@"C"))) = -350color(red)(cancel(color(black)("g"))) * 4.200color(red)(cancel(color(black)("J")))/(color(red)(cancel(color(black)("g"))) color(red)(cancel(color(black)(""^@"C")))) * (color(blue)(T_"f") - 80)color(red)(cancel(color(black)(""^@"C")))#

#190 * color(blue)(T_"f") - 3800 = -1470 * color(blue)(T_"f") + 117600#

#1660 * color(blue)(T_"f") = 121400 implies color(blue)(T_"f") = 121400/1660 = 73.13^@"C"#

Now, you *should* round this off to one sig fig, the number of sig figs you have for the mass of milk, but I'll leave it rounded off to two sig figs

#T_"f" = color(green)(73^@"C")#

Indeed, the result confirms our prediction - tea's higher *specific heat* and *bigger mass* ensured that the final temperature of the mixture will be much closer to its initial temperature.