Question

suppose you mix 100.0 g of water at 26.3 °c with 75.0 g of water at 75.0 °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 conservation of energy, \(q_{lost}=q_{gained}\). The heat - transfer formula is \(q = mc\Delta T\), where \(m\) is the mass, \(c\) is the specific heat capacity (for water \(c = 4.18\space J/g^{\circ}C\), and since it is the same substance \(c\) cancels out), \(\Delta T=T - T_0\). Let the final temperature be \(T\). The mass of the first sample \(m_1 = 100.0\space g\) and its initial temperature \(T_{01}=26.3^{\circ}C\), the mass of the second sample \(m_2 = 75.0\space g\) and its initial temperature \(T_{02}=75.0^{\circ}C\). Then \(m_1c(T - T_{01})=m_2c(T_{02}-T)\).

Step2: Simplify the equation

Cancel out \(c\) (because \(c
eq0\)) from both sides of the equation \(m_1(T - T_{01})=m_2(T_{02}-T)\). Expand the equation: \(m_1T-m_1T_{01}=m_2T_{02}-m_2T\).

Step3: Rearrange the terms to solve for \(T\)

Group the terms with \(T\) on one side: \(m_1T + m_2T=m_2T_{02}+m_1T_{01}\). Factor out \(T\): \(T(m_1 + m_2)=m_2T_{02}+m_1T_{01}\). Then \(T=\frac{m_2T_{02}+m_1T_{01}}{m_1 + m_2}\).

Step4: Substitute the values

Substitute \(m_1 = 100.0\space g\), \(T_{01}=26.3^{\circ}C\), \(m_2 = 75.0\space g\), \(T_{02}=75.0^{\circ}C\) into the formula:

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Answer:

\(47.2^{\circ}C\)