Question

current attempt in progress
three portions of the same liquid are mixed in a container that prevents the exchange of heat with the environment. portion a has a mass m and a temperature of 95.0 °c, portion b also has a mass m but a temperature of 85.0 °c, and portion c has a mass m_c and a temperature of 38.0 °c. what must be the mass of portion c so that the final temperature t_f of the three-portion mixture is t_f = 55.0 °c? express your answer in terms of m; for example, m_c = 2.20 m.

m_c = number ← units ← *m

Explanation:

Step1: Apply heat transfer formula

According to the principle of heat transfer \(Q = cm\Delta T\) (where \(c\) is the specific heat capacity, \(m\) is the mass, and \(\Delta T\) is the temperature change). Since there is no heat exchange with the environment \(Q_{A}+Q_{B}+Q_{C}=0\).
For portion \(A\): \(Q_{A}=cm(m)(T_{f}-T_{A})\) (here \(T_{A} = 95.0^{\circ}C\), \(T_{f}=55.0^{\circ}C\)), so \(Q_{A}=cm(m)(55 - 95)=- 40cm(m)\)
For portion \(B\): \(Q_{B}=cm(m)(T_{f}-T_{B})\) (here \(T_{B}=85.0^{\circ}C\)), so \(Q_{B}=cm(m)(55 - 85)=-30cm(m)\)
For portion \(C\): \(Q_{C}=cm(m_{C})(T_{f}-T_{C})\) (here \(T_{C}=38.0^{\circ}C\)), so \(Q_{C}=cm(m_{C})(55 - 38)=17cm(m_{C})\)

Step2: Solve for \(m_{C}\)

Substitute \(Q_{A}\), \(Q_{B}\), and \(Q_{C}\) into \(Q_{A}+Q_{B}+Q_{C}=0\)
\(-40cm(m)-30cm(m)+17cm(m_{C}) = 0\)
First, combine the terms with \(m\): \(-70cm(m)+17cm(m_{C})=0\)
Then, divide both sides of the equation by \(cm\) (since \(c
eq0\)): \(- 70m+17m_{C}=0\)
Solve for \(m_{C}\): \(m_{C}=\frac{70}{17}m\approx4.12m\)

Answer:

\(m_{C} = 4.12m\)