How much heat must be added to 1-kg of 0 degrees Celsius water to turn it into water at 50 degrees Celsius? If the water had been ice at 0 degrees Celsius, would that increase, decrease, or not change the amount of energy required?
To calculate the amount of heat required to raise the temperature of water, you can use the specific heat capacity of water. The specific heat capacity of water is approximately 4.18 J/g°C.
First, convert the mass of water from kilograms to grams:
1 kg = 1000 g
The heat equation is given by:
Q = mcΔT
Where:
Q is the heat required (in joules),
m is the mass of the substance (in grams),
c is the specific heat capacity (in joules per gram per degree Celsius),
ΔT is the change in temperature (in degrees Celsius).
Let's calculate the amount of heat required to raise the temperature of 1 kg of water from 0 to 50 degrees Celsius:
m = 1000 g
c = 4.18 J/g°C
ΔT = (50 - 0) °C = 50 °C
Q = mcΔT
Q = 1000 g * 4.18 J/g°C * 50 °C
Q = 209,000 J
Therefore, it would require 209,000 joules of heat to raise the temperature of 1 kg of water from 0 to 50 degrees Celsius.
Now, let's consider the second part of your question. If the water had been ice at 0 degrees Celsius, it would undergo a phase change from solid to liquid during the heating process. This phase change requires additional energy, known as the heat of fusion.
The heat of fusion for water is approximately 334 J/g (joules per gram). Therefore, if the water was in the solid state (ice), an additional amount of energy would be required to melt the ice before raising its temperature.
To calculate the total amount of energy required to turn ice at 0 degrees Celsius into water at 50 degrees Celsius, we need to consider two steps:
1. The energy required to melt the ice (change the solid into a liquid)
2. The energy required to raise the temperature of the liquid water from 0 to 50 degrees Celsius
For step 1 (melting ice):
m = 1000 g (mass of 1 kg of ice)
Q1 = mc (heat of fusion)
Q1 = 1000 g * 334 J/g
Q1 = 334,000 J
For step 2 (raising temperature):
m = 1000 g (mass of 1 kg of liquid water)
c = 4.18 J/g°C
ΔT = 50 °C - 0 °C = 50 °C
Q2 = mcΔT
Q2 = 1000 g * 4.18 J/g°C * 50 °C
Q2 = 209,000 J
Total energy required = Q1 + Q2
Total energy required = 334,000 J + 209,000 J
Total energy required = 543,000 J
Therefore, if the water had been ice at 0 degrees Celsius, it would require a total of 543,000 joules of energy to turn it into water at 50 degrees Celsius.