If 4.168 kJ of heat is added to a calorimeter containing 75.40 g of water, the temperature of the water and the calorimeter increases from 24.58°C to 35.82°C. Calculate the heat capacity of the calorimeter (in J/°C). The specific heat of water is 4.184
Well, well, well, looks like we've got a hot situation here! Calculating the heat capacity of the calorimeter is a piece of cake, so let's dive right in, shall we?
First things first, we need to find the change in temperature of the water and the calorimeter. We have:
Initial temperature (T1) = 24.58°C
Final temperature (T2) = 35.82°C
So, ΔT = T2 - T1 = 35.82°C - 24.58°C = 11.24°C. Got it? Great!
Now, we know that heat capacity (C) is defined as the amount of heat (Q) required to change the temperature of an object by 1 degree Celsius. In this case, we need to find the heat capacity of the calorimeter (C_calorimeter).
Using the equation Q = m * c * ΔT, where Q is the amount of heat, m is the mass, c is the specific heat capacity of the substance, and ΔT is the change in temperature, we can solve for C_calorimeter.
Let me crunch those numbers for you...
Q = 4.168 kJ = 4.168 * 10^3 J
m = 75.40 g
ΔT = 11.24°C
c = 4.184 J/g°C
Plugging in the values, we have:
4.168 * 10^3 J = (75.40 g + C_calorimeter) * 4.184 J/g°C * 11.24°C
Simplifying further...
4.168 * 10^3 J = (75.40 g + C_calorimeter) * 47.0256 J
Now, divide both sides by the constant 47.0256 J...
(4.168 * 10^3) / 47.0256 = 75.40 g + C_calorimeter
Solving for C_calorimeter...
C_calorimeter = (4.168 * 10^3) / 47.0256 - 75.40 g
And there you have it! The heat capacity of the calorimeter is (4.168 * 10^3) / 47.0256 - 75.40 g J/°C. Just make sure you double-check your units!
If you have any more questions, feel free to drop them my way. I'm always here to bring some laughter to your day!
To calculate the heat capacity of the calorimeter, we can use the formula:
q = mcΔT
Where:
q is the heat added to the system
m is the mass of the substance
c is the specific heat capacity of the substance
ΔT is the change in temperature
In this case, the substance is a combination of water and the calorimeter, so we need to consider both of their heat capacities.
First, let's calculate the heat capacity of the water:
q_water = m_water * c_water * ΔT_water
Given:
m_water = 75.40 g
c_water = 4.184 J/g°C
ΔT_water = (35.82°C - 24.58°C)
q_water = 75.40 g * 4.184 J/g°C * (35.82°C - 24.58°C)
Now, let's calculate the heat capacity of the calorimeter:
q_calorimeter = m_calorimeter * c_calorimeter * ΔT_calorimeter
Since we want to find the heat capacity of the calorimeter, we need to rewrite the formula as:
C_calorimeter = q_calorimeter / ΔT_calorimeter
To find q_calorimeter, we can subtract q_water from the total heat added to the system:
q_calorimeter = q_total - q_water
q_total = 4.168 kJ = 4.168 * 1000 J
Now we can find C_calorimeter:
C_calorimeter = (q_total - q_water) / ΔT_calorimeter
Substituting the values we found earlier:
C_calorimeter = (4.168 * 1000 J - 75.40 g * 4.184 J/g°C * (35.82°C - 24.58°C)) / (35.82°C - 24.58°C)
To calculate the heat capacity of the calorimeter, you need to use the formula:
Q = mcΔT
where:
Q is the heat added to the calorimeter
m is the mass of the water
c is the specific heat capacity of water
ΔT is the temperature change of the water in the calorimeter
First, convert the mass of water from grams to kilograms:
m = 75.40 g = 0.07540 kg
Next, calculate the temperature change of the water:
ΔT = final temperature - initial temperature
ΔT = 35.82°C - 24.58°C = 11.24°C
Now, calculate the heat added to the calorimeter using the given value:
Q = 4.168 kJ = 4.168 * 10^3 J
Substitute the values into the formula:
4.168 * 10^3 J = (0.07540 kg) * (4.184 J/g°C) * (11.24 °C) + C
Simplify the equation:
4168 J = (0.07540 kg) * (4.184 J/g°C) * (11.24 °C) + C
Multiply and divide to eliminate units:
4168 J = 4.184 * (0.07540 kg) * (11.24) + C
Now, solve for C:
4168 J = 4.184 * 0.845896 kg + C
4168 J = 3.537826624 J + C
Subtract 3.537826624 J from both sides:
4168 J - 3.537826624 J = C
C = 4168 J - 3.537826624 J
C ≈ 4164.46 J
Therefore, the heat capacity of the calorimeter is approximately 4164.46 J/°C.
q = Ccal*(Tfinal-Tinitial) + [mass H2O x specific heat H2O x (Tfinal-Tinitial)]
Solve for Ccal
Watch the units. If you use 4.184 J/g*C for H2O and you want Ccal in J/C, then you must substitute q in J, not kJ.