A copper calorimeter of mass 150g was half filled with water of mass 300g and temperature 0^C. 5g of ice at 0^C was added to the content, later some steam was passed into the mixture and the temperature rose by 20^C. Calculate the quantity of steam added. (Specific latent heat of vaporisation of steam = 2.26* 10^6JKg^-1 specific heat of copper=400JKg-1K-1)
Solution for: A copper calorimeter of mass 150g was half filled with water of mass 300g and temperature 0^C. 5g of ice at 0^C was added to the content, later some steam was passed into the mixture and the temperature rose by 20^C. Calculate the quantity of steam added. (Specific latent heat of vaporisation of steam = 2.26* 10^6JKg^-1 specific heat of copper=400JKg-1K-1)
Please can you help with this question 🙏
A copper calorimeter of mass 150g was half filled with water of mass 300g and temperature 5°c.50g of alluminium at 35°c was added to the content which in the temperature of the water to 20°c.calculate the temperature change of the calorimeter.(s.h.c.of copper=400j/kgk,s.h.c.of water=4200j/kgk and s.h.c.of alluminium=900j/kgk)
To solve this problem, we can use the principle of heat exchange, which states that the heat gained by the water and the ice is equal to the heat lost by the steam and the copper calorimeter.
First, let's calculate the initial heat of the water and the ice using the equation:
Q = mcΔT
where:
Q is the heat energy
m is the mass
c is the specific heat capacity
ΔT is the change in temperature
For the water:
m_water = 300g
c_water = 4200 J/kg°C (specific heat capacity of water)
ΔT_water = 20°C (change in temperature of the water)
Q_water = m_water * c_water * ΔT_water
Q_water = 300g * 4200 J/kg°C * 20°C
= 2,520,000 J
For the ice:
m_ice = 5g
c_ice = 2100 J/kg°C (specific heat capacity of ice)
ΔT_ice = 20°C (change in temperature of the ice, which remains at 0°C)
Q_ice = m_ice * c_ice * ΔT_ice
Q_ice = 5g * 2100 J/kg°C * 20°C
= 210,000 J
Now, let's calculate the heat required to melt the ice using the equation:
Q_melt = m_ice * L_f
where:
m_ice is the mass of the ice
L_f is the specific latent heat of fusion (for ice, it is 334,000 J/kg)
Q_melt = 5g * 334,000 J/kg
= 1,670,000 J
The total initial heat absorbed is the sum of Q_water and Q_ice:
Q_initial = Q_water + Q_ice + Q_melt
= 2,520,000 J + 210,000 J + 1,670,000 J
= 4,400,000 J
Next, let's calculate the heat gained by the copper calorimeter:
Q_copper = m_copper * c_copper * ΔT_water
where:
m_copper = 150g (mass of the copper calorimeter)
c_copper = 400 J/kg°C (specific heat capacity of copper)
ΔT_copper = 20°C (change in temperature of the copper calorimeter)
Q_copper = 150g * 400 J/kg°C * 20°C
= 1,200,000 J
Finally, the heat lost by the steam is equal to the sum of Q_initial and Q_copper. Since the heat lost by steam is equal, the heat gained is also equal:
Q_lost = Q_initial + Q_copper
Therefore, the quantity of steam added can be calculated by dividing the total heat lost by the specific latent heat of vaporization of steam:
Quantity of steam added = Q_lost / L_v
where:
L_v is the specific latent heat of vaporization of steam (2.26 * 10^6 J/kg)
Let's calculate:
Quantity of steam added = (4,400,000 J + 1,200,000 J) / 2.26 * 10^6 J/kg
Quantity of steam added = 5,600,000 J / 2.26 * 10^6 J/kg
= 2.48 kg (approx.)
Therefore, approximately 2.48 kg of steam was added to the mixture.