When 1.375g CaO is added to 50 mL water at 25.0 degree C, the temperature of the water increases to 34.6 degree C. Given the specific heat is 4.184 j/g*degree C, how much heat is absorbed by the calorimeter?
q = mass H2O x specific heat H2O x (Tfinal-Tinitial) is the heat absorbed.
Then q/1.375 is heat absorbed per gram CaO and that x molar mass CaO is heat absorbed per mol CaO. I think those questions would follow although you didn't post them yet.
To find out how much heat is absorbed by the calorimeter, we need to calculate the heat gained by the water and then subtract it from the total heat gained by the system.
To calculate the heat gained by the water, we can use the formula:
Q = m * c * ΔT
where:
Q is the heat gained by the water
m is the mass of the water
c is the specific heat of water
ΔT is the change in temperature
In this case, the mass of water is given as 50 mL, so we need to convert it to grams by considering the density of water, which is approximately 1 g/mL.
mass of water = volume of water * density = 50 mL * 1 g/mL = 50 g
The specific heat of water, c, is given as 4.184 J/g*°C.
The change in temperature, ΔT, can be calculated by subtracting the initial temperature from the final temperature:
ΔT = final temperature - initial temperature = 34.6 °C - 25.0 °C = 9.6 °C
Now, we can calculate the heat gained by the water:
Q = 50 g * 4.184 J/g*°C * 9.6 °C = 2013.12 J
This means that 2013.12 Joules of heat are gained by the water.
To find out how much heat is absorbed by the calorimeter, we need to subtract the heat gained by the water from the total heat gained by the system. In this case, the total heat gained is the same as the heat gained by the CaO.
The heat gained by the CaO can be calculated using the formula:
Q = m * c * ΔT
where:
Q is the heat gained by the CaO
m is the mass of the CaO
c is the specific heat of CaO (assumed to be the same as water's specific heat for simplicity)
ΔT is the change in temperature of the CaO (assumed to be the same as water's change in temperature for simplicity)
The mass of CaO is given as 1.375 g.
The change in temperature of the CaO can be calculated by subtracting the initial temperature from the final temperature:
ΔT = final temperature - initial temperature = 34.6 °C - 25.0 °C = 9.6 °C
Now we can calculate the heat gained by the CaO:
Q = 1.375 g * 4.184 J/g*°C * 9.6 °C = 53.57 J
So, the heat absorbed by the calorimeter is 53.57 J.