I'm stuck on this problem. "You mix 100 ml 60 degree C water with 100ml 20 degree C water. Find the final temperature" Changing 100 ml to .1kg, I set .1*4186*(60-final temp.) + .1*4186*(final temp-20)=0 but my t's cancel out. I know it would work if I set my first half of equation to (Final temp-60) but writing it this way doesn't make sense to me since the final temp. is somewhere between 20 and 60. Please help!!
Your should take final-initial temperature on both sides. On left hand side, u r taking initial minus final whereas on right hand side u r taking final minus initial which is wrong
To solve this problem, we can apply the principle of energy conservation in the form of heat transfer. The heat gained by the colder water must be equal to the heat lost by the hotter water. We can use the equation:
\(Q_{\text{gained}} = Q_{\text{lost}}\)
Where \(Q_{\text{gained}}\) is the heat gained by the colder water, and \(Q_{\text{lost}}\) is the heat lost by the hotter water.
Now, let's calculate these quantities:
\(Q_{\text{gained}} = m_{\text{colder}} \cdot c_{\text{water}} \cdot (T_f - T_{\text{colder}})\)
Where \(m_{\text{colder}}\) is the mass of the colder water, \(c_{\text{water}}\) is the specific heat capacity of water, \(T_f\) is the final temperature, and \(T_{\text{colder}}\) is the initial temperature of the colder water.
\(Q_{\text{lost}} = m_{\text{hotter}} \cdot c_{\text{water}} \cdot (T_{\text{hotter}} - T_f)\)
Where \(m_{\text{hotter}}\) is the mass of the hotter water, and \(T_{\text{hotter}}\) is the initial temperature of the hotter water.
Since in this case, both masses are 0.1 kg and the specific heat capacity of water \(c_{\text{water}}\) is 4186 J/kg·°C, we can simplify the equations:
\(418.6 \cdot (T_f - 20) = 418.6 \cdot (60 - T_f)\)
Now, let's solve for \(T_f\):
\(418.6T_f - 8372 = 25016 - 418.6T_f\)
\(8372 = 837.2T_f\)
\(T_f = \frac{8372}{837.2}\)
\(T_f \approx 10 \, \text{°C}\)
Therefore, the final temperature is approximately 10°C.