A copper Calorimeter of mass 150g,was half-filled with water of mass 300g and a temperature of 0°C.5g of ice of mass 0°C was added to the content, later some quantity of steam was passed into the mixture and the temperature rose by 20°C. Calculate the quantity of steam added.
A copper calorimeter of mass 150g,was heated filled with water of mass 300g at temperature 0 degree celsius,5g of ice at 0 degree was added to the content,later some quantity of steam was passed into the mixture and the temperature rose by 20 degree celsius.calculate the quantity of steam added.(specific latent heat of vapourisation of steam =2.26*10^-16J/Kg) and (specific latent heat of copper=400J/Kg/K)?
Calculate
To solve this problem, we need to understand the concepts of heat transfer and the specific heat capacity.
The calorimeter is made of copper, so we need to consider the specific heat capacity of copper. The specific heat capacity of copper is 0.39 J/g°C.
The water in the calorimeter is initially at 0°C. To calculate the heat gained by the water when it rises to a final temperature of 20°C, we can use the following equation:
Q = m × c × ΔT
Where:
Q is the heat gained (in Joules),
m is the mass of the water (in grams),
c is the specific heat capacity of water (which is 4.18 J/g°C), and
ΔT is the change in temperature (in °C).
Next, we need to calculate the heat gained by the calorimeter. The calorimeter has a mass of 150g and a specific heat capacity of 0.39 J/g°C. We use the same equation:
Q = m × c × ΔT
Where:
Q is the heat gained (in Joules),
m is the mass of the calorimeter (in grams),
c is the specific heat capacity of the calorimeter (which is 0.39 J/g°C), and
ΔT is the change in temperature (in °C).
Now, let's calculate the heat absorbed by the ice to reach its final temperature of 0°C. The ice has a mass of 5g and a specific heat capacity of 2.09 J/g°C. We use the same equation:
Q = m × c × ΔT
Where:
Q is the heat gained (in Joules),
m is the mass of the ice (in grams),
c is the specific heat capacity of ice (which is 2.09 J/g°C), and
ΔT is the change in temperature (in °C).
Finally, we need to calculate the heat necessary to convert the ice at 0°C to steam at 100°C. This is called the heat of vaporization, and for water, it is 540 cal/g (or 2260 J/g). The heat required can be calculated using the formula:
Q = m × ΔH
Where:
Q is the heat gained (in Joules),
m is the mass of the substance (in grams), and
ΔH is the heat of vaporization (in Joules/gram).
Now, let's calculate the heat absorbed by the system (water + calorimeter + ice) to reach a final temperature of 20°C. We can add up the heat gained by each component:
Qsystem = Qwater + Qcalorimeter + Qice + Qsteam
Next, we need to calculate the heat absorbed by the steam to raise the temperature by 20°C. We use the same equation:
Qsteam = msteam × csteam × ΔT
Where:
Qsteam is the heat gained by the steam (in Joules),
msteam is the mass of the steam (in grams),
csteam is the specific heat capacity of steam (which is 2.03 J/g°C), and
ΔT is the change in temperature (in °C).
Now, let's calculate the total mass of the system:
Total mass = mass of water + mass of calorimeter + mass of ice + mass of steam
To find the quantity of steam added, we use the equation:
Quantity of steam = mass of steam added - mass of ice added
By substituting the values into the equations and performing the calculations, we can find the quantity of steam added.