What is the temperature change when 9.8 g of KNO3( with 34.9 kJ/mol enthaply of solution) is dissolved in 150 g of water? Assume that the specific heat capacity is 4.18 J/K*g. Answer in celcius.
So far I have the Q= ^T*m*s but i don't know what numbers go where and what not.
^T= delta T
KNO3(s) + H2O + heat ==> KNO3(aq)
q = mass x specific heat x deltaT
change 9.8 g KNO3 to mols and calculate q from the 34.9 kJ/mol. Change to J.
Plug that number in for q, mass is mass of water (150 g), specific heat is that of water (4.184 J/g*C) and delta T is the unknown. solve for delta T. Post your work if you get stuck.
To calculate the temperature change when 9.8 g of KNO3 is dissolved in 150 g of water, you can follow these steps:
Step 1: Convert the mass of KNO3 into moles. You can do this by dividing the mass by the molar mass of KNO3. The molar mass of KNO3 is 101.1 g/mol (39.1 g/mol for K, 14.0 g/mol for N, and 16.0 g/mol for each of the three O atoms).
9.8 g KNO3 / 101.1 g/mol = 0.097 mol KNO3
Step 2: Calculate the heat change using the given enthalpy of solution of KNO3.
q = ΔH * n
where q is the heat change, ΔH is the enthalpy of solution (34.9 kJ/mol), and n is the number of moles of KNO3.
q = 34.9 kJ/mol * 0.097 mol = 3.38 kJ
Note: Since specific heat capacity is given in units of J/K*g, we need to convert the heat change from kJ to J (1 kJ = 1000 J).
q = 3.38 kJ * 1000 J/kJ = 3380 J
Step 3: Now we can use the heat equation q = mcΔT to calculate the temperature change (ΔT).
q = m * c * ΔT
where q is the heat change, m is the mass of water, c is the specific heat capacity of water, and ΔT is the temperature change.
Substituting the given values:
3380 J = 150 g * 4.18 J/g*C * ΔT
Step 4: Solve for ΔT.
Divide both sides of the equation by (150 g * 4.18 J/g*C):
ΔT = 3380 J / (150 g * 4.18 J/g*C)
ΔT = 0.453 °C
Therefore, the temperature change when 9.8 g of KNO3 is dissolved in 150 g of water is approximately 0.453 °C.