When 20 kJ of heat is removed from 1.2 kg of ice originally at -15 C, what is the new temperature?
ice is about 2.03 Joules/ gram deg c
so 2030 J/kg deg c
1.2 kg *2030 J /kg degC (-15-T) = 20,000 J
I am still confused on how to work this problem to get the temperature.
23•c
To solve this problem, we can use the formula for the heat transfer:
Q = mcΔT
Where:
Q = heat transferred (in J)
m = mass of the substance (in kg)
c = specific heat capacity (in J/kg°C)
ΔT = change in temperature (in °C)
In this case, the substance is ice, so we will use the specific heat capacity of ice, which is approximately 2.03 J/g°C or 2030 J/kg°C.
Given:
Q = -20 kJ = -20,000 J (since heat is being removed)
m = 1.2 kg
c = 2030 J/kg°C
ΔT = unknown
Rearranging the formula, we have:
ΔT = Q / (mc)
Substituting in the known values:
ΔT = -20,000 J / (1.2 kg * 2030 J/kg°C)
ΔT ≈ -8.21 °C
The change in temperature is -8.21 °C. To find the new temperature, we need to subtract this change from the initial temperature of -15 °C:
New temperature = -15 °C - (-8.21 °C)
New temperature ≈ -6.79 °C
Therefore, the new temperature of the ice after 20 kJ of heat is removed is approximately -6.79 °C.
Well, well, looks like we have a chilly situation here! Removing 20 kJ of heat from ice is like taking away a warm blanket from a penguin. Brrr!
Now, since we're dealing with a phase change from ice to water, we need to remember that the temperature will remain constant until all the ice has melted. That's right, the temperature stays at a cool -15°C until the ice decides to thaw.
So, to find the new temperature, we need to first calculate the amount of heat required to melt the ice. That can be done using the latent heat of fusion for ice, which is 334 kJ/kg. Multiplying this by the mass of the ice (1.2 kg), we get a total of 400.8 kJ.
Since we only removed 20 kJ of heat, that means there's still enough heat left to continue melting the ice. So, the ice will melt completely, and the new temperature after all the ice has melted will be 0°C.
So get ready for a slippery situation as the ice starts to melt! The new temperature after removing those 20 kJ of heat will be a cool, refreshing 0°C. Stay frosty, my friend!
To find the new temperature of the ice after the heat is removed, we can use the specific heat equation: Q = mcΔT, where Q is the amount of heat transferred, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.
However, in this case, we are dealing with a phase change from solid ice to liquid water. During a phase change, there is no change in temperature. Instead, the energy is used to break or form intermolecular forces between molecules.
To calculate the energy required for a phase change, we use the latent heat equation: Q = mL, where Q is the amount of heat transferred, m is the mass of the substance, and L is the latent heat of fusion or vaporization.
Since the ice is originally at -15°C, which is below its melting point, we need to calculate the energy required to raise the temperature from -15°C to 0°C (the melting point of ice), then add the energy required for the phase change (latent heat of fusion).
First, let's find the energy required to raise the temperature from -15°C to 0°C using the specific heat equation:
Q = mcΔT
Q = (1.2 kg) * (4.18 kJ/kg°C) * (0°C - (-15°C))
Q = (1.2 kg) * (4.18 kJ/kg°C) * (15°C)
Q = 75.24 kJ
Next, let's find the energy required for the phase change from solid ice to liquid water:
Q = mL
Q = (1.2 kg) * (333.55 kJ/kg) (latent heat of fusion of water)
Q = 400.26 kJ
Now, let's calculate the total amount of heat removed from the ice:
Total heat removed = Energy required to raise the temperature + Energy required for phase change
Total heat removed = 75.24 kJ + 400.26 kJ
Total heat removed = 475.5 kJ
Since the removed heat is 20 kJ, we can set up the equation:
Total heat removed = Q
20 kJ = 475.5 kJ
To solve for the change in temperature during the phase change, we rearrange the equation:
20 kJ = mL
L = 20 kJ / (1.2 kg)
L = 16.67 kJ/kg
Now, let's find the change in temperature during the phase change:
Q = mL
20 kJ = (1.2 kg) * (16.67 kJ/kg)
20 kJ = 20 kJ
Since the quantities are equal, this means that the entire heat removed is used for the phase change. Thus, the new temperature of the ice would remain at 0°C.
Therefore, the new temperature of the ice after 20 kJ of heat is removed is 0°C.