Suppose you mix 20.1 g of water at 66.2°C with 44.6 g of water at 35.7°C in an insulated cup. What is the maximum temperature of the solution after mixing?
3 answers
[mass1 water x specific heat water x (Tfinal-Tinitial)] + [mass2 water x specific heat x (Tfinal-Tinitla)] = 0
simplified energy balance:
Q,absorbed + Q,lost = 0 , or
Q,absorbed = -Q,lost
where Q = m*c(Tf - Ti)
in which Q=heat in J, m=mass in kg, c=specific heat capacity (which has a definite value for various substances) in J/(kg*C), T=temperature in Kelvin
*Tf=final temp (the same for both and is usually the unknown)
*Ti=initial temp
since in this problem, it involves only one type of substance (which is water), the c will be canceled:
Q,absorbed = -Q,lost
m1*c*(Tf - Ti) = -m2*c*(Tf - Ti)
m1*(Tf - Ti) = -m2*(Tf - Ti)
then substitute the given:
20.1*(T2 - 66.2) = -44.6*(T2 - 35.7)
simplify and solve for Tf. (units in degree Celsius)
hope this helps. :)
Q,absorbed + Q,lost = 0 , or
Q,absorbed = -Q,lost
where Q = m*c(Tf - Ti)
in which Q=heat in J, m=mass in kg, c=specific heat capacity (which has a definite value for various substances) in J/(kg*C), T=temperature in Kelvin
*Tf=final temp (the same for both and is usually the unknown)
*Ti=initial temp
since in this problem, it involves only one type of substance (which is water), the c will be canceled:
Q,absorbed = -Q,lost
m1*c*(Tf - Ti) = -m2*c*(Tf - Ti)
m1*(Tf - Ti) = -m2*(Tf - Ti)
then substitute the given:
20.1*(T2 - 66.2) = -44.6*(T2 - 35.7)
simplify and solve for Tf. (units in degree Celsius)
hope this helps. :)
thanks :)