50 mL of water at 48.9EC were mixed with 50 mL of water at 23.8EC in a calorimeter also at 23.8EC. The final temperature was 33.1EC. Assuming that neither the density of water nor its specific heat capacity change with temperature, calculate the total heat capacity of the calorimeter. (density of water = 1.00 g mL-1, specific heat capacity = 4.18 J g-1 K-1)
heat lost by water + heat gained by calorimeter = 0
[(mass H2O x specific heat H2O x (Tfinal-Tinitial)] + [(Ccal*(Tfinal-Tinitial)] = 0
The ionic compund sometimes called yellow uranium is used to produce colored glazes for ceramics. It is 7.252% sodium, 75.084% uranium, 17.664% oxygen?
a. What is the empirical formula for this compound?
b. This compound has a molecular mass of 1268.06 g/mol, what is the molecuslar compound for this formula?
It is better to post your own question instead of riding piggy back on another post. I've already answered your question above.
To calculate the total heat capacity of the calorimeter, we can use the principle of conservation of energy.
The amount of heat gained by the water at 48.9 °C is given by the formula:
q1 = m1 * c1 * ΔT1
where:
- q1 represents the heat gained by the water at 48.9 °C
- m1 represents the mass of water at 48.9 °C
- c1 represents the specific heat capacity of water
- ΔT1 represents the change in temperature for the water at 48.9 °C
Plugging in the known values:
m1 = 50 mL = 50 g (since the density of water is 1.00 g/mL)
ΔT1 = final temperature - initial temperature = 33.1 °C - 48.9 °C = -15.8 °C (note: the negative sign indicates a temperature drop)
Using the specific heat capacity of water, c1 = 4.18 J/g·K, we can calculate the heat gained by the water at 48.9 °C:
q1 = 50 g * 4.18 J/g·K * -15.8 °C = -3126.8 J
Similarly, we can calculate the heat gained by the water at 23.8 °C:
q2 = m2 * c2 * ΔT2
where:
- q2 represents the heat gained by the water at 23.8 °C
- m2 represents the mass of water at 23.8 °C
- c2 represents the specific heat capacity of water
- ΔT2 represents the change in temperature for the water at 23.8 °C
Plugging in the known values:
m2 = 50 mL = 50 g
ΔT2 = final temperature - initial temperature = 33.1 °C - 23.8 °C = 9.3 °C
Using the specific heat capacity of water, c2 = 4.18 J/g·K, we can calculate the heat gained by the water at 23.8 °C:
q2 = 50 g * 4.18 J/g·K * 9.3 °C = 1948.5 J
Based on the principle of conservation of energy, the total heat released by the water at 48.9 °C must be equal to the total heat absorbed by the water at 23.8 °C and the calorimeter:
q1 = q2 + qcal
where:
- qcal represents the heat absorbed by the calorimeter
Substituting the calculated values:
-3126.8 J = 1948.5 J + qcal
To isolate qcal, we rearrange the equation:
qcal = -3126.8 J - 1948.5 J
qcal = -5075.3 J
Therefore, the total heat capacity of the calorimeter is -5075.3 J. Notice that the negative sign indicates that the calorimeter absorbs heat.