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?

[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. :)

thanks :)

To find the maximum temperature of the solution after mixing, we can use the principle of energy conservation.

In this problem, the heat lost by the hot water (initially at 66.2°C) is equal to the heat gained by the cold water (initially at 35.7°C). We can use the equation:

Q_hot + Q_cold = 0

where Q_hot is the heat lost by the hot water and Q_cold is the heat gained by the cold water.

To calculate the heat lost or gained, we can use the equation:

Q = m * c * ΔT

where Q is the heat transferred, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.

For the hot water:
Q_hot = m_hot * c_water * ΔT_hot

For the cold water:
Q_cold = m_cold * c_water * ΔT_cold

Since the final temperature of the solution is the same, we have:
ΔT_hot = ΔT_final
ΔT_cold = ΔT_final

Now let's calculate the heat lost by the hot water:
Q_hot = 20.1 g * c_water * (66.2°C - ΔT_final)

And the heat gained by the cold water:
Q_cold = 44.6 g * c_water * (ΔT_final - 35.7°C)

Since the total heat lost is equal to the total heat gained, we can set up the equation:

20.1 g * c_water * (66.2°C - ΔT_final) + 44.6 g * c_water * (ΔT_final - 35.7°C) = 0

Simplifying and rearranging the equation, we get:

(20.1 g + 44.6 g) * c_water * ΔT_final = (20.1 g * 66.2°C) + (44.6 g * 35.7°C)

Now we can solve for ΔT_final:

ΔT_final = [(20.1 g * 66.2°C) + (44.6 g * 35.7°C)] / (20.1 g + 44.6 g)

Calculating this expression will give us the maximum temperature of the solution after mixing.