New problem. Any assistance is greatly appreciated.
22.5 ml of ethanol (density=0.789 g/ml) initially at 7.7 deg. C is mixed with 31.6 ml of water (density=1.0 g/ml) initially at 27.1 deg. C in an insulated beaker. Assuming that no heat is lost, what is the final temperature of the mixture. Express your answer using two significant figures.
22.5ml EtOH 22.5ml x 0.789g/1ml= 17.75 g
31.6ml water 31.6ml x 1g/ml=31.6g
total mass= 49.4 g
This is where I am at a loss. There is no q given. q= mass x SHC x delta temp.
SHC ethanol 2.42 J/g*C
SHC water 4.184 J/g*C
Maybe I am making this more complicated than it should be. Any direction now is appreciated. Since there is no heat lost, do I assume that q=1 or do I just subtract the two temps. 27.1*C - 7.7*C=
The sum of heats gained is zero (some lose heat).
heat gained by ethanol+heatgained by water=0
massethanol*Ceth*(Tf-7.7)+masswater*Cwater*(Tf-27.1)=0
solve for Tf
To solve this problem, you need to consider the heat transfer between the ethanol and water during the mixing process. Since the system is insulated and no heat is lost, the heat gained by water will be equal to the heat lost by ethanol.
The equation that relates heat transfer to the change in temperature is given by:
q = m * C * delta T
Where:
q = heat transfer in joules (J)
m = mass of the substance in grams (g)
C = specific heat capacity of the substance in J/(g*C)
delta T = change in temperature in degrees Celsius (°C)
Let's calculate the heat transfer for both the ethanol and water:
Heat transfer for ethanol:
m_ethanol = 17.75 g (mass of ethanol)
C_ethanol = 2.42 J/(g°C) (specific heat capacity of ethanol)
delta T_ethanol = T_final - T_initial = T_final - 7.7°C
Heat transfer for water:
m_water = 31.6 g (mass of water)
C_water = 4.184 J/(g°C) (specific heat capacity of water)
delta T_water = T_final - T_initial = T_final - 27.1°C
Since the heat gained by water is equal to the heat lost by ethanol, we can set up the equation:
m_water * C_water * delta T_water = m_ethanol * C_ethanol * delta T_ethanol
Now, substitute the known values:
31.6 * 4.184 * (T_final - 27.1) = 17.75 * 2.42 * (T_final - 7.7)
Solve this equation to find the value of T_final, which represents the final temperature of the mixture.
Rearranging and simplifying the equation:
133 * (T_final - 27.1) = 42.925 * (T_final - 7.7)
133T_final - 133 * 27.1 = 42.925T_final - 42.925 * 7.7
Now, isolate T_final:
133T_final - 3603.3 = 42.925T_final - 331.8625
90.075T_final = 3271.4375
T_final = 36.29°C (rounded to two significant figures)
Therefore, the final temperature of the mixture is approximately 36.29°C.