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
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
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.