Question

suppose you mix 100.0 g of water at 23.1 °c with 75.0 g of water at 77.7 °c. what will be the final temperature of the mixed water, in °c? type answer:

Explanation:

Step1: Set up the 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 specific heat capacity \(c\) of water is the same for both samples, so it cancels out. The equation becomes \(m_1c(T - T_1)=m_2c(T_2 - T)\), and then \(m_1(T - T_1)=m_2(T_2 - T)\)

Step2: Substitute the values

Given \(m_1 = 100.0\space g\), \(T_1=23.1^{\circ}C\), \(m_2 = 75.0\space g\), \(T_2 = 77.7^{\circ}C\). Substitute into \(m_1(T - T_1)=m_2(T_2 - T)\):

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Answer:

\(46.5^{\circ}C\)