Answer:
67.80 mL
Explanation:
The heat q is given by the formula:
[tex]q = mC_p(T_2-T_1)[/tex]
Since heat released by the coffee = - heat absorbed by the milk; Then :
[tex]q_{coffee}= -q_{milk}[/tex]
[tex]m_{coffee}C_p(T_2-T_1)_{coffee} = -m_{milk}C_p(T_2-T_1)_{milk}[/tex]
From the question;
given that the heat capacity of milk & coffee are equal; The density are also said to be equal:
The temperature difference for coffee is as follows:
[tex](T_2-T_1)_{coffee} = 60.00^ ^0 }}C - 80.00 ^ ^ 0 }}C = -20.00 ^ ^ 0 }}C[/tex]
Temperature difference for milk is :
[tex](T_2-T_1)_{milk} = 60.00^ ^0 }}C - 1.00 ^ ^ 0 }}C = 59.00 ^ ^ 0 }}C[/tex]
We all know that : Density [tex]\rho[/tex] = mass (m) / volume (v)
then m = v × [tex]\rho[/tex]
So, we can say that : [tex]m_{coffee}C_p(T_2-T_1)_{coffee} = -m_{milk}C_p(T_2-T_1)_{milk}[/tex] can now be re-written as:
[tex]v_{coffee}*\rho _{coffee}* (T_2-T_1)_{coffee} = -v_{milk}*\rho_{milk}*(T_2-T_1)_{milk}[/tex]
[tex]v_{coffee}* (T_2-T_1)_{coffee} = -v_{milk}*(T_2-T_1)_{milk}[/tex]
Replacing the values; we have :
[tex]200 \ ml * (-20.0^0 C)= -v_{milk}*(59^0 C)[/tex]
[tex]v_{milk} = \frac{200 \ ml *(-20.0^0 C)}{-59^0 \ C}[/tex]
[tex]v_{milk} = 67.80 \ mL[/tex]
Therefore, the amount of milk required to reach the required temperature for coffee is 67.80 mL