Respuesta :
Answer:
The answer is "36.76 ml "
Explanation:
In the given question, Its basic understanding to also be hired seems to be that coffee heat = heat that the milk absorbs
The following relationship defines heat:
[tex]\to q = m C_{p}( T_2 - T_1)[/tex]
Its updated equation is:
[tex]\to q_{\ coffee} = - q_\ milk} \\\\ \to m_{\ milk} C_{p} (T_2 - T_1)_{\ coffee} = - m_{ \ milk} C_{p} (T_2 - T_1)_{\ milk} ..........(a)[/tex]
The following information is provided to you as per the question:
1) Milk and coffee have the same heat power
2) Difference in coffee temperature is:
[tex]\to (T_2 - T_1)_{\ coffee} = 75.0^{\circ} \ \ C-85.0^{\circ} \ \ C[/tex]
[tex]= -10.0 ^{\circ} \ \ C[/tex]
3) The difference in milk temperature:
[tex]\to (T_2 - T_1)_{ \ milk} = 75.0^{\circ} \ \ C- 7.0^{\circ} \ \ C[/tex]
[tex]= 68.0 ^{\circ} \ \ C[/tex]
4) Milk and coffee density are equal therefore, weight and volume are equal and have same ratio, that can be shown as follows:
[tex]\to \ Density (\rho) =\frac{ \ Mass (m)}{ \ Volume (V)} \\\\\to m= V \times \rho \\\\[/tex]
[tex]\to m_{\ coffee} \times (T_2 - T_1)_{ \ coffee}= -m_{\ milk} \times (T_2 - T_1)_{milk} \\\\\to V_{\ coffee} \times \rho_{\ coffee} \times (T_2 - T_1)_{\ coffee} = - V_{\ milk} \times \rho_{\ milk} \times (T_2 - T_1)_{\ milk} \\\\\to V_{\ coffee} \times (T_2 - T_1)_{\ coffee} = - V_{\ milk} \times (T_2 - T_1)_{\ milk}.............(b)[/tex]
by replace the equation values (b) for calculating milk volumes as shown as follows:
[tex]\to V_{\ coffee} \times (T_2 - T_1)_{\ coffee} = - V_{\ milk} \times (T_2 - T_1)_{\ milk}\\\\\to 250 ml \times (-10 .0^{\circ} \ \ C) = - V_{\ milk} \times (68.0^{\circ} \ \ C)\\\\\to V_{\ milk}= \frac{250 ml \times (-10 .0^{\circ} \ \ C) }{- (68.0^{\circ} \ \ C)}[/tex]
[tex]= 36.76 \ ml[/tex]
