solution at 18°C is mixed with 3.00 dm^3 of the saturated solution at 70°C, what is the final temperature of the mixture?
To solve this problem, we can use the principle of conservation of energy. The heat gained by the 18°C solution will be equal to the heat lost by the 70°C solution. The heat gained or lost can be calculated using the equation:
q = m * c * ΔT
Where:
q is the heat gained or lost (in Joules)
m is the mass of the solution (in grams)
c is the specific heat capacity of water (4.18 J/g°C)
ΔT is the change in temperature (in °C)
First, we need to calculate the mass of the solutions in grams. Since the density of water is 1 g/cm^3, the mass of the solutions in grams will be equal to their volumes in cm^3:
Mass of 1.33 dm^3 of water at 18°C = 1.33 * 1000 = 1330 g
Mass of 4.5 dm^3 of saturated solution at 18°C = 4.5 * 1000 = 4500 g
Next, let's calculate the heat gained by the 18°C solution:
q1 = m1 * c * ΔT1
= 1330 * 4.18 * (T_final - 18)
Now, let's calculate the heat lost by the 70°C solution:
q2 = m2 * c * ΔT2
= 4500 * 4.18 * (70 - T_final)
Since q1 = q2 (according to the principle of conservation of energy), we can set them equal and solve for the final temperature:
1330 * 4.18 * (T_final - 18) = 4500 * 4.18 * (70 - T_final)
Dividing both sides of the equation by 4.18 and simplifying:
1330 * (T_final - 18) = 4500 * (70 - T_final)
1330T_final - 23880 = 4500 * 70 - 4500T_final
1330T_final + 4500T_final = 4500 * 70 + 23880
5830T_final = 315000
T_final = 54.00°C
Therefore, the final temperature of the mixture is 54.00°C.
1•33dm^3 of water at 70°c are saturated by 2•25 mole of lead( ll) trioxonitrate pb( NO ^3) ^2 and 1•33dm^3 of water at 18°c are saturated by 0•53 mole of the same salt if 4•50 dm^3 or The saturated
1 answer