Question

1. consider mixing 0.12 kg of 26°c water with 0.08 kg of 50°c water. assume the specific heat of water is 4186 j/kg°c. the final temperature of the mixture is

(a) 58.9 °c
(b) 28.9 °c
(c) 76.9 °c
(d) 38.0 °c
(d) none of the above

Explanation:

Step1: Set up heat transfer equation

According to the principle of heat transfer \(Q = mc\Delta T\), and \(Q_{lost}=Q_{gained}\). Let the final temperature be \(T\). The heat lost by the hot water is \(Q_{lost}=m_2c(T_2 - T)\), and the heat gained by the cold water is \(Q_{gained}=m_1c(T - T_1)\). So \(m_2c(T_2 - T)=m_1c(T - T_1)\). Since \(c\) (specific heat of water) cancels out on both sides.

Step2: Substitute values

Given \(m_1 = 0.12\space kg\), \(T_1=26^{\circ}C\), \(m_2 = 0.08\space kg\), \(T_2 = 50^{\circ}C\). Substitute into \(m_2(T_2 - T)=m_1(T - T_1)\), we get \(0.08\times(50 - T)=0.12\times(T - 26)\).

Step3: Expand and solve for \(T\)

Expand the equation: \(4-0.08T=0.12T - 3.12\).
Combine like - terms: \(4 + 3.12=0.12T+0.08T\).
\(7.12 = 0.2T\).
Solve for \(T\): \(T=\frac{7.12}{0.2}=35.6^{\circ}C\).

Answer:

(e) none of the above