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 (d) 40.0 °c (e) none of the above

Explanation:

Step1: Apply the principle of heat transfer

According to the principle of heat transfer \(Q = mc\Delta T\), and in an isolated system \(Q_{lost}=Q_{gained}\). Let the final temperature be \(T\). For the first problem, the heat gained by the cooler water \(Q_1=m_1c(T - T_1)\) and the heat lost by the warmer water \(Q_2=m_2c(T_2 - T)\). Since \(Q_1 = Q_2\), we have \(m_1c(T - T_1)=m_2c(T_2 - T)\). Canceling out \(c\) (specific heat of water), we get \(m_1(T - T_1)=m_2(T_2 - T)\).
Substituting \(m_1 = 0.12\space kg\), \(T_1=26^{\circ}C\), \(m_2 = 0.08\space kg\), \(T_2 = 50^{\circ}C\) into the equation:

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Since \(35.6^{\circ}C\) is not among the options (a)-(d) for the first problem, the answer for the first problem is (e) none of the above.

For the second problem, assume the specific heat of water \(c_w=4186\space J/kg^{\circ}C\) and specific heat of soya - bean oil \(c_{soya}\approx1970\space J/kg^{\circ}C\). Using \(Q_{lost}=Q_{gained}\), \(m_wc_w(T - T_{w1})=m_{soya}c_{soya}(T_{soya1}-T)\)
Substituting \(m_w = 0.15\space kg\), \(T_{w1}=30^{\circ}C\), \(m_{soya}=0.055\space kg\), \(T_{soya1} = 50^{\circ}C\), \(c_w = 4186\space J/kg^{\circ}C\), \(c_{soya}=1970\space J/kg^{\circ}C\)

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Answer:

For the first problem: (e) none of the above.
For the second problem: (a) \(32.94^{\circ}C\)