What would be the final temp if you mixed a liter of 20 degreesCelsius with 2 li

Question

What would be the final temp if you mixed a liter of 20 degreesCelsius with 2 liters of 40 degrees Celsius.. Pretty sure theanswer is 35 degrees, but is there a mathematical way to find that?Thanks

Explanation / Answer

Let system 1 be the 20 C and 1 Liter sample Let system 2 be the 40 C and 2 Liter sample mass of system 1 = m1 = 0.5*m2 , where m2 = mass of system 2. (youknow 2 is twice as large as 1 because of the volumes) Heat absorbed by system 1 = Heat released by system 2 m1 *cp1 *(T1_final - T1_initial) = m2*cp2*(T2_initial -T2_final).      note T2_initial >T2_final, hence the order of subtraction cp1 = cp2 = heat capacity per mass. You can cancel these out m1*(T1_final - T1_initial) = m2*(T2_initial - T2_final) m1*(T1_final - T1_initial) = 2*m1*(T2_initial - T2_final) (T1_final - T1_initial) = 2*(T2_initial -T2_final)               Note that T1_final = T2_final = T_final (equilibrium) T_final - T1_initial = 2*T2_initial _ 2T_final 3T_final = 2*T2_initial + T1_initial = 2*40 + 20 = 100 T_final = 100/3 = 33.3 Celsius

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