3. consider mixing 0.12 kg of 26°c water with 0.08 kg of 50°c water. assume the specific heat of water is…

3. 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

3. 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

Answer

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