Calculate the heat energy required to change 0.1kg of ice at 0oc to water boiling at 100oc (SPC of water = 4200, SLH of ice =336000)?
heat to melt the ice: Hf*.1kg
Heat to heat water to 100C= .1kg*4200j/kg*100
calculate, then add, watch units. Specifically, numbers such as you gave without units are totally meaningless
To calculate the heat energy required, we need to consider the phase changes and the temperature changes involved. Let's break down the calculation into three parts:
1. Heat energy required to melt the ice:
The heat energy required to change the ice at 0°C to water at 0°C can be calculated using the formula:
Q1 = mass × specific latent heat of fusion (SLH)
Here, mass = 0.1 kg and the specific latent heat of fusion (SLH) is given as 336,000 J/kg.
Q1 = 0.1 kg × 336,000 J/kg
Q1 = 33,600 J
So, 33,600 J of heat energy is required to melt the ice.
2. Heat energy required to raise the temperature of the water from 0°C to 100°C:
The heat energy required to raise the temperature of the water can be calculated using the formula:
Q2 = mass × specific heat capacity (SPC) × ΔT
Here, mass = 0.1 kg, specific heat capacity (SPC) of water is given as 4,200 J/kg°C, and ΔT is the change in temperature, which is 100°C - 0°C = 100°C.
Q2 = 0.1 kg × 4,200 J/kg°C × 100°C
Q2 = 42,000 J/°C × 100°C
Q2 = 4,200,000 J
So, 4,200,000 J of heat energy is required to raise the temperature of the water.
3. Heat energy required to vaporize the water:
The heat energy required to vaporize water at 100°C can be calculated using the formula:
Q3 = mass × specific latent heat of vaporization (SLV)
Here, mass = 0.1 kg and specific latent heat of vaporization (SLV) is the same as the specific latent heat of fusion for water, which is given as 336,000 J/kg.
Q3 = 0.1 kg × 336,000 J/kg
Q3 = 33,600 J
So, 33,600 J of heat energy is required to vaporize the water.
To get the total heat energy required, we can add up all the calculated values:
Total heat energy = Q1 + Q2 + Q3
Total heat energy = 33,600 J + 4,200,000 J + 33,600 J
Total heat energy = 4,267,200 J
Therefore, the heat energy required to change 0.1 kg of ice at 0°C to water boiling at 100°C is 4,267,200 J.