Question

Question #da1ce

Answer 1

`T_"f" = 73^@"C"`

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 `"1.00-g"` sample of that substance to change its temperature by `1^@"C"` or by `"1 K"`.

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 `1^@"C"` change in its temperature.

The equation you're going to use looks like this

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

`q` - heat absorbed/lost
`m` - the mass of the sample
`c` - the specific heat of the substance
`DeltaT` - the change in temperature, defined as 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 `color(blue)(T_"f")`. You can say that the 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 `color(purple)((1))` and solve for `color(blue)(T_"f")`

`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.