What will be the final temperature if 1.9 x 10^2 kJ of heat is added to 0.5 kg of ice at 0 deg. C?
It takes 500g*335 J/g*C = 167,500 J to melt all the ice. If you add 190,000 J, you melt all the ice and have 22,500 J left to heat the water.
Use the relation
22,500 J = M C *(delta T)
to determine the temperature rise, delta T.
C = 4.18 J/g*C M = 500 g
To find the final temperature when heat is added to ice, we need to consider the processes involved in the phase change from solid to liquid and then to a higher temperature.
First, we need to calculate the heat required to change the ice into water at its melting point (0°C). This can be done using the formula:
Q = m * L
Where:
- Q is the heat energy in joules
- m is the mass of the substance in kilograms
- L is the latent heat of fusion for the substance (the amount of heat required to change a unit mass of the substance from solid to liquid)
For ice, the latent heat of fusion is approximately 334 kJ/kg.
Q1 = m * L = 0.5 kg * 334 kJ/kg = 167 kJ
The heat energy required to convert the ice to water is 167 kJ.
Next, we calculate the heat energy required to raise the temperature of the water from 0°C to the final temperature. We use the specific heat capacity (c) of water, which represents the amount of heat required to raise the temperature of 1 kg of the substance by 1°C. For water, the specific heat capacity is approximately 4.18 kJ/kg°C.
Q2 = m * c * ΔT
Where:
- Q2 is the heat energy in joules
- m is the mass of the substance in kilograms
- c is the specific heat capacity of the substance
- ΔT is the change in temperature in degrees Celsius
In this case, the change in temperature is the final temperature minus the starting temperature (0°C). Let's assume the final temperature is Tf.
Q2 = 0.5 kg * 4.18 kJ/kg°C * (Tf - 0°C)
Finally, we add up the heats from both processes to find the total heat energy added to the system.
Q total = Q1 + Q2 = 167 kJ + 0.5 kg * 4.18 kJ/kg°C * (Tf - 0°C)
Now, by substituting the known values, we can solve for the final temperature (Tf).
Given:
Q total = 1.9 x 10^2 kJ
1.9 x 10^2 kJ = 167 kJ + 0.5 kg * 4.18 kJ/kg°C * (Tf - 0°C)
Simplifying the equation:
1.9 x 10^2 kJ = 167 kJ + 2.09 kJ/kg°C * Tf
1.9 x 10^2 kJ - 167 kJ = 2.09 kJ/kg°C * Tf
33 x 10^2 kJ = 2.09 kJ/kg°C * Tf
33 x 10^2 kJ / (2.09 kJ/kg°C) = Tf
Tf ≈ 158.37°C
Hence, the final temperature of the system after adding 1.9 x 10^2 kJ of heat will be approximately 158.37°C.