How much thermal energy is absorbed by 1.00x10^2g of ice at
-20.0°C to become water at 0.0°C?
Two stages to go through.
q1 = heat to move ice from -20 C to 0.0 C
q2 = heat to convert ice @ 0.0 C to liquid H2O @ 0.0 C
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q1 = mass ice x specific heat ice x (Tfinal - Tinitial)
q2 = mass ice x heat fusion ice
Total heat = q1 + q2
Post your work if you get stuck.
Question , why do we have to do heat the liquid after melting the ice if it’s not specified in the question
Like some quesiton it says to make hot chocolate so we know we have to hit the liquid but why are we heating it in this question
Thank you
To calculate the thermal energy absorbed by the ice, we need to determine the amount of heat required to change the temperature of the ice from -20.0°C to 0.0°C and then the heat required to change the ice at 0.0°C into water at the same temperature.
The heat (Q) absorbed or released by a substance can be calculated using the formula:
Q = m * c * ΔT
Where:
Q = heat absorbed or released (in Joules)
m = mass of the substance (in grams)
c = specific heat capacity of the substance (in J/g°C)
ΔT = change in temperature (in °C)
1. Calculate the heat required to change the temperature of the ice from -20.0°C to 0.0°C:
Q1 = m * c * ΔT1
Given:
m = 1.00x10^2g (mass of the ice)
c = 2.09 J/g°C (specific heat capacity of ice)
ΔT1 = (0.0°C - (-20.0°C)) = 20.0°C
Q1 = (1.00x10^2g) * (2.09 J/g°C) * (20.0°C)
Q1 = 4.18x10^3 J
Therefore, the heat required to change the temperature of the ice from -20.0°C to 0.0°C is 4.18x10^3 Joules.
2. Calculate the heat required to change the ice at 0.0°C into water at the same temperature:
Q2 = m * ΔHf
Where:
ΔHf = heat of fusion for ice (in J/g)
Given:
m = 1.00x10^2g (mass of the ice)
ΔHf = 333.5 J/g (heat of fusion for ice)
Q2 = (1.00x10^2g) * (333.5 J/g)
Q2 = 3.335x10^4 J
Therefore, the heat required to change the ice at 0.0°C into water at the same temperature is 3.335x10^4 Joules.
Finally, we can find the total thermal energy absorbed by adding the two calculated amounts:
Total heat absorbed = Q1 + Q2
Total heat absorbed = 4.18x10^3 J + 3.335x10^4 J
Total heat absorbed = 3.757x10^4 J
Therefore, the total thermal energy absorbed by 1.00x10^2g of ice at -20.0°C to become water at 0.0°C is 3.757x10^4 Joules.
To determine the amount of thermal energy absorbed by a substance, you need to use the heat equation, which is:
Q = m * c * ΔT
where:
Q is the thermal energy absorbed or released (in Joules),
m is the mass of the substance (in grams),
c is the specific heat capacity of the substance (in J/g·°C),
ΔT is the change in temperature (in °C).
In this case, you want to find the thermal energy absorbed by 1.00x10^2g of ice when it undergoes a temperature change from -20.0°C to 0.0°C.
First, let's find the change in temperature (ΔT):
ΔT = final temperature - initial temperature
ΔT = 0.0°C - (-20.0°C)
ΔT = 20.0°C
Next, you need the specific heat capacity of ice. The specific heat capacity of ice is 2.09 J/g·°C.
Now, you can substitute the values into the heat equation:
Q = m * c * ΔT
Q = (1.00x10^2g) * (2.09 J/g·°C) * (20.0°C)
Q = 41,800 J
Therefore, the thermal energy absorbed by 1.00x10^2g of ice when it changes from -20.0°C to 0.0°C is 41,800 Joules.