A 3.90×10^2kg cube of ice at an initial temperature of -18.0degreesC is placed in 0.480kg of water at 41.0degreesC in an insulated container of negligible mass.
Calculate the change in entropy of the system.
I found the equilibrium temperature to be at 304.9K, then used
change in S = mass*c*ln(T2/T1) where T2 is the final temperature and T1 is the initial temperature.
But my answer was wrong.
There are TWO initial temperatures. You can not use the equation you used, except for the cooling of the liquid water. When ice is melting (if that ever happens here), the temperature stays the same and the entropy gain is Qin/273K
Are you sure you copied the numbers correctly? There is nearly 1000 times more ice than liquid water initially, according to the numbers you gave. If that is true, you would heat the ice but not melt any.
Rethink the equilibrium temperature. Then calculate the entropy gain of the ice and subtract the entropy loss of the water.
To calculate the change in entropy of the system, you can use the formula:
ΔS = ΔSice + ΔSwater
where ΔSice is the change in entropy of the ice cube and ΔSwater is the change in entropy of the water.
To calculate the change in entropy of the ice cube, you can use the formula:
ΔSice = mi * specific heat capacity of ice * ln(Tf/Ti)
where mi is the mass of the ice cube, specific heat capacity of ice is approximately 2.09 J/g°C, Tf is the final temperature (in Kelvin), and Ti is the initial temperature (in Kelvin).
First, convert the given mass of the ice cube from kg to grams:
mi = 3.90 × 10^2 kg * 1000 g/kg = 3.90 × 10^5 g
Convert the initial temperature of the ice cube from -18.0 °C to Kelvin:
Ti = -18.0 °C + 273.15 = 255.15 K
Next, calculate the mass of the water and the final temperature of the system. Since the system is insulated, the heat lost by the water is equal to the heat gained by the ice cube:
Qlost = Qgained
where Qlost is the heat lost by the water and Qgained is the heat gained by the ice cube.
To calculate the heat lost by the water, you can use the formula:
Qlost = mw * specific heat capacity of water * (Tf - Ti)
where mw is the mass of the water, specific heat capacity of water is approximately 4.18 J/g°C, Tf is the final temperature (in Kelvin), and Ti is the initial temperature (in Kelvin).
Given:
mw = 0.480 kg * 1000 g/kg = 480 g
Ti = 41.0 °C + 273.15 = 314.15 K
Rearrange the formula to solve for Tf:
Tf = Qlost / (mw * specific heat capacity of water) + Ti
Substitute the values into the equation and solve for Tf.
Once you have Tf, use the formula for ΔSice to calculate the change in entropy of the ice cube, and ΔSwater to calculate the change in entropy of the water.
Finally, calculate the total change in entropy of the system by summing up the individual changes:
ΔS = ΔSice + ΔSwater