How many grams of ice at 0°C and 101.3 kPa (standard pressure) could be melted by the addition of 0.80 kJ of heat?
Note that the process involved here is the phase change of ice, from solid (ice) to liquid (water). Thus, the heat involved is latent heat, not sensible heat.
The latent heat of fusion of ice is approximately 334 J/g. You can look for this value in textbooks or google.
Then we use the formula,
Q = m*Lf
where
m = mass (g)
Lf = latent heat of fusion
Substituting (note that 1 kJ = 1000 J),
0.80 kJ * (1000 J / 1 kJ) = m * 334 J/g
m = 800 / 334
m = 2.40 g
hope this helps~ `u`
Well, that seems like a pretty cool question! To determine the number of grams of ice that can be melted, we need to consider the heat required to melt the ice. The heat of fusion for ice is 334 J/g. So, if we divide the given amount of heat (0.80 kJ or 800 J) by the heat of fusion (334 J/g), we can find out how many grams of ice can be melted.
So, 800 J divided by 334 J/g gives us approximately 2.40 grams of ice. But hey, don't get too frozen in those calculations! Just remember that this is an approximation, as there may be some slight variations in the heat of fusion depending on the specific conditions. Ice to meet you!
To determine the number of grams of ice that could be melted by the addition of 0.80 kJ of heat, we need to use the specific heat formula.
The specific heat of ice is 2.09 J/g°C, and the heat required to melt ice is known as the heat of fusion and is 334 J/g.
First, let's convert the heat energy from kJ to J:
0.80 kJ = 0.80 × 1000 J = 800 J
Next, we can use the following formula:
Q = m × c × ΔT
Where:
Q = heat energy (J)
m = mass (grams)
c = specific heat (J/g°C)
ΔT = change in temperature (°C)
Since the ice is initially at 0°C and is being melted, there is no change in temperature. Therefore, ΔT = 0.
800 J = m × 2.09 J/g°C × 0
This simplifies to:
800 J = 0
Since 0 J is not equal to 800 J, it means that the heat energy provided (800 J) is not enough to cause a change in the temperature or melt the ice.
Therefore, 0 grams of ice could be melted by the addition of 0.80 kJ of heat at 0°C and 101.3 kPa.
To answer this question, we need to know the specific heat capacity of ice. The specific heat capacity of a substance is the amount of heat required to raise the temperature of one gram of that substance by one degree Celsius.
The specific heat capacity of ice is 2.09 J/g°C (joules per gram degree Celsius).
First, we convert the heat energy from kilojoules to joules:
0.80 kJ x 1000 J/kJ = 800 J.
Next, we use the equation:
q = m * c * ΔT
where:
q = heat energy (in joules)
m = mass of the substance (in grams)
c = specific heat capacity (in J/g°C)
ΔT = change in temperature (in °C)
In this case, the change in temperature is from 0°C (the initial temperature of the ice) to the melting point of ice, which is also 0°C. Therefore, ΔT = 0 - 0 = 0°C.
Plugging in the values we have:
800 J = m * 2.09 J/g°C * 0°C
Now we can solve for the mass of the ice, m:
800 J = m * 0 J
m = 800 J / 0 J
m = undefined
Since the change in temperature is 0°C, no ice will be melted by the addition of 0.80 kJ of heat. The formula above gives us an undefined value because there is no change in temperature, and therefore no change in state of the substance.