How much ice at 0 degrees celsius can be melted with 1kg of steam at 100 degrees celsius, assuming no loss of energy to the surroundings? Use specific latent heat values. The specific heat of water is 4200 J kg-1 K-1
Well, let's do some math! We need to find out how much ice at 0 degrees Celsius can be melted by 1 kg of steam at 100 degrees Celsius.
First, we need to find out how much energy is required to melt the ice. The specific latent heat of fusion for ice is 334,000 J/kg. This means that it takes 334,000 J of energy to melt 1 kg of ice at 0 degrees Celsius.
Now, let's find out how much energy can be released by 1 kg of steam at 100 degrees Celsius. The specific heat capacity of water is 4200 J/kg K, which means it takes 4200 J of energy to raise the temperature of 1 kg of water by 1 degree Celsius.
To raise the temperature of 1 kg of water from 100 degrees Celsius to its boiling point, 100 degrees Celsius, we need:
4200 J/kg K x 100 K = 420,000 J
So, 1 kg of steam at 100 degrees Celsius can release 420,000 J of energy.
To find out how much ice can be melted, we divide the energy released by the energy required to melt the ice:
420,000 J / 334,000 J/kg = 1.257 kg
Therefore, approximately 1.257 kg of ice at 0 degrees Celsius can be melted by 1 kg of steam at 100 degrees Celsius, assuming no loss of energy to the surroundings.
Now that's a cool calculation, isn't it?
To calculate the amount of ice that can be melted, we need to consider the energy transfer between the steam and the ice.
First, let's calculate the energy required to raise the ice from 0°C to its melting point, which is also 0°C.
Q1 = mass of ice × specific heat of ice × temperature change
= 1 kg × 2100 J kg-1 K-1 × 0°C
= 0 J
The temperature of the ice will rise from 0°C to 0°C, so no energy is required.
Next, let's calculate the energy required to melt the ice at 0°C.
Q2 = mass of ice × specific latent heat of fusion
= 1 kg × 334,000 J kg-1
= 334,000 J
Finally, let's calculate the energy released by the steam as it cools down from 100°C to 0°C.
Q3 = mass of steam × specific heat of water × temperature change
= 1 kg × 4200 J kg-1 K-1 × (100°C - 0°C)
= 420,000 J
Since the system is assumed to have no energy loss, the energy released by the steam is equal to the energy required to melt the ice.
Q3 = Q2
420,000 J = 334,000 J
Therefore, 1 kg of steam at 100°C can melt 334,000 J of ice at 0°C.
Note: To convert the energy from joules (J) to calories (cal), divide by 4.184.
To determine how much ice at 0 degrees Celsius can be melted with 1 kg of steam at 100 degrees Celsius, we need to consider the energy exchange that occurs during the phase change.
First, let's calculate the energy required to heat the ice from 0 degrees Celsius to its melting point. The specific heat capacity of water is given as 4200 J/kg/K, meaning it takes 4200 Joules of energy to raise the temperature of 1 kg of water by 1 degree Celsius.
Since we need to heat the ice from 0 degrees to its melting point at 0 degrees, it will require (0 - 0) * 4200 J = 0 Joules of energy.
Next, we need to calculate the energy required to melt the ice at its melting point. The latent heat of fusion of ice is 334,000 J/kg. This means it takes 334,000 Joules of energy to convert 1 kg of ice at 0 degrees Celsius to 1 kg of water at 0 degrees Celsius.
Now, let's calculate the energy that can be provided by 1 kg of steam at 100 degrees Celsius. The specific latent heat for changing steam at 100 degrees Celsius to water at 100 degrees Celsius is 2,260,000 J/kg. This means it takes 2,260,000 Joules of energy to convert 1 kg of steam at 100 degrees Celsius to 1 kg of water at 100 degrees Celsius.
Since we know that energy is conserved and there is no loss of energy to the surroundings, the energy required to heat and melt the ice should be equal to the energy provided by the steam:
Energy to heat ice + Energy to melt ice = Energy from steam
0 J + 334,000 J = 2,260,000 J
Therefore, the amount of ice that can be melted with 1 kg of steam at 100 degrees Celsius, assuming no loss of energy to the surroundings, is 334,000 J / 2,260,000 J = 0.1478 kg of ice.
So, 1 kg of steam at 100 degrees Celsius can melt approximately 0.1478 kg of ice at 0 degrees Celsius.