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 \(Q_{lost}=Q_{gained}\). Let the final temperature be \(T\). For the cold water (\(m_1 = 0.12\space kg\), \(T_1=26^{\circ}C\)) and hot water (\(m_2 = 0.08\space kg\), \(T_2 = 50^{\circ}C\)), \(m_1c(T - T_1)=m_2c(T_2 - T)\). Since \(c\) (specific heat of water) cancels out on both sides.

Step2: Solve for \(T\)

Substitute the values: \(0.12(T - 26)=0.08(50 - T)\)
Expand: \(0.12T-3.12 = 4-0.08T\)
Combine like - terms: \(0.12T + 0.08T=4 + 3.12\)
\(0.2T=7.12\)
\(T=\frac{7.12}{0.2}=35.6^{\circ}C\)

Answer:

E. none of the above