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 assume the specific heat capacity of water is 4186 j/kg°c. 4. consider mixing 0.15 kg of 30°c water with 0.055 kg of 50°c soya bean oil. the final temperature of the mixture is (a) 32.94°c (b) 45.9°c (c) 58.9°c

Explanation:

Step1: Apply heat transfer formula

According to the principle of heat transfer \(Q = mc\Delta T\), and in the case of no heat loss \(Q_{lost}=Q_{gain}\). Let the final temperature be \(T\). For water, \(m_1 = 0.12\space kg\), \(T_1=26^{\circ}C\), \(m_2 = 0.08\space kg\), \(T_2 = 50^{\circ}C\), \(c = 4186\space J/kg^{\circ}C\). Then \(m_1c(T - T_1)=m_2c(T_2 - T)\).

Step2: Simplify the equation

Cancel out \(c\) (since \(c\) is the same for water) from both sides of the equation \(m_1(T - T_1)=m_2(T_2 - T)\). Substitute \(m_1 = 0.12\), \(T_1 = 26\), \(m_2=0.08\), \(T_2 = 50\) into the equation: \(0.12(T - 26)=0.08(50 - T)\).

Step3: Expand and solve for \(T\)

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

Answer:

E. none of the above