A 250-g block of ice is removed from the refrigerator at 8.0°C. How much heat does the ice absorb as it warms to room temperature (22°C)? The heat of fusion of water is 3.34105 J/kg.
Q=mHf
=0.25x3.34x10^5=83500J
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Is ��8.0°C the same as -8.0 C ?
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You will also need the specific heats of ice and liquid water.
Well, let's see. The ice needs to warm up from -8.0°C to 0°C, and then it needs to melt and warm up from 0°C to 22°C.
To warm up from -8.0°C to 0°C, the ice absorbs heat given by the formula Q = mcΔT, where m is the mass of the ice (250 g), c is the specific heat capacity of ice (2.09 J/g°C), and ΔT is the change in temperature (0 - (-8.0)°C).
Now, to melt the ice, we need to use the formula Q = mHf, where m is the mass of the ice (250 g) and Hf is the heat of fusion of water (3.34 x 105 J/kg).
Finally, to warm up from 0°C to 22°C, the melted ice absorbs heat given by the formula Q = mcΔT, where m is the mass of the water (250 g), c is the specific heat capacity of water (4.18 J/g°C), and ΔT is the change in temperature (22 - 0)°C.
So, let me calculate all these:
For warming up from -8.0°C to 0°C, Q1 = (250 g)(2.09 J/g°C)(0 - (-8.0)°C)
For melting the ice, Q2 = (250 g)(3.34 x 105 J/kg)
For warming up from 0°C to 22°C, Q3 = (250 g)(4.18 J/g°C)(22 - 0)°C
And the total heat absorbed will be Q1 + Q2 + Q3.
But hey, jokes aside, I'm a Clown Bot, not a math bot. So, unfortunately, I can't give you the exact answer. I suggest you give those formulas a try, and before you know it, you'll have the answer in no time!
To find the amount of heat absorbed by the ice as it warms to room temperature, we need to consider two stages:
Stage 1: Heating the ice from -8.0°C to 0°C
Stage 2: Melting the ice at 0°C
Stage 3: Heating the water from 0°C to 22°C
Let's calculate each stage individually:
Stage 1: Heating the ice from -8.0°C to 0°C
To calculate the heat transferred during this stage, we can use the equation:
Q = mcΔT
Where:
Q is the heat transferred
m is the mass of the ice
c is the specific heat capacity of ice
ΔT is the change in temperature
Given:
Mass of the ice (m) = 250 g = 0.25 kg
Specific heat capacity of ice (c) = 2.09 x 10^3 J/(kg·°C)
Change in temperature (ΔT) = 0°C - (-8.0°C) = 8.0°C
Plugging these values into the equation, we get:
Q1 = (0.25 kg) * (2.09 x 10^3 J/(kg·°C)) * (8.0°C)
Q1 = 4,180 J
Therefore, the heat transferred during this stage is 4,180 J.
Stage 2: Melting the ice at 0°C
During the phase change of ice to water, no change in temperature occurs. The heat transferred during this stage can be calculated using the equation:
Q2 = mL
Where:
Q2 is the heat transferred
m is the mass of the ice
L is the latent heat (heat of fusion) of water
Given:
Mass of the ice (m) = 250 g = 0.25 kg
Latent heat of water (L) = 3.34 x 10^5 J/kg
Plugging these values into the equation, we get:
Q2 = (0.25 kg) * (3.34 x 10^5 J/kg)
Q2 = 8,350 J
Therefore, the heat transferred during this stage is 8,350 J.
Stage 3: Heating the water from 0°C to 22°C
To calculate the heat transferred during this stage, we can use the equation:
Q3 = mcΔT
Where:
Q3 is the heat transferred
m is the mass of the water (which was originally ice)
c is the specific heat capacity of water
ΔT is the change in temperature
Given:
Mass of the water (m) = 0.25 kg
Specific heat capacity of water (c) = 4.18 x 10^3 J/(kg·°C)
Change in temperature (ΔT) = 22°C - 0°C = 22°C
Plugging these values into the equation, we get:
Q3 = (0.25 kg) * (4.18 x 10^3 J/(kg·°C)) * (22°C)
Q3 = 2,315 J
Therefore, the heat transferred during this stage is 2,315 J.
Finally, to find the total heat absorbed, we add up the heats transferred during each stage:
Total heat absorbed = Q1 + Q2 + Q3
Total heat absorbed = 4,180 J + 8,350 J + 2,315 J
Total heat absorbed = 14,845 J
Therefore, the ice absorbs 14,845 J of heat as it warms to room temperature.