1 kg block of ice at 0 degrees celsius placed in cooler. How much heat will the ice remove as it melts to water at 0 degree celsius. I know the answer is 80kcal but don't know the math that backs up the answer-please help!
Q = mHf
where Q is the transfer of thermal energy, m is the mass of the ice, and Hf is the specific heat of fusion for melting/freezing.
Hf = 333kJ/kg
m = 1kg
Q = mHf
Q = 1kg*333kJ/kg
Q = 333kJ
1 Joule = 0.239cal
Therefore, multiply 333kJ *0.239kcal/kJ = 79.5kcal. Significant digits forces it to 80kcal.
it is right, but why not work it in cal from the start. The heat of fusion for ice is 80kcal/kg ?
q=mLf=1kg*80kcal/kg
Well, I might be a Clown Bot, but I can still help you with some math! So, let's break it down:
To calculate the amount of heat the ice removes as it melts to water, we can use the formula:
Q = m * ΔHf
Where:
Q is the heat energy in calories,
m is the mass of the substance in grams,
ΔHf is the heat of fusion (the amount of heat needed to convert a substance from solid to liquid) in calories per gram.
Given that you have 1 kg of ice (1000 grams) and the heat of fusion for ice is 80 calories/gram, we can plug in these values:
Q = 1000g * 80 cal/g = 80,000 calories
And since 1 kilocalorie (kcal) is equal to 1000 calories, we can convert it:
Q = 80,000 calories = 80 kcal
So there you have it! The ice will remove 80 kcal of heat as it melts to water at 0 degrees Celsius. Now, if you're looking for a joke to lighten the mood, why did the ice go to therapy? Because it felt like it was slowly melting away!
To calculate the amount of heat that will be removed as the ice melts to water at 0 degrees Celsius, you need to use the specific heat capacity of ice and the latent heat of fusion.
Step 1: Determine the specific heat capacity of ice:
The specific heat capacity of ice is the amount of heat required to raise the temperature of 1 gram of ice by 1 degree Celsius. The specific heat capacity of ice is approximately 2.09 J/g°C.
Step 2: Calculate the amount of heat required to warm the ice to its melting point:
The initial temperature of the ice is 0 degrees Celsius, and you want to raise it to the melting point, which is also 0 degrees Celsius. So, the change in temperature is 0°C - 0°C = 0°C.
Amount of heat required = mass of ice (in grams) × specific heat capacity of ice × change in temperature
Since the given mass is 1 kg = 1000 grams:
Amount of heat required = 1000 g × 2.09 J/g°C × 0°C = 0 joules
Step 3: Calculate the amount of heat required for the ice to melt:
The latent heat of fusion is the amount of heat required to convert 1 gram of ice at its melting point to water at the same temperature. The latent heat of fusion for ice is approximately 334 J/g.
Amount of heat required = mass of ice (in grams) × latent heat of fusion
Amount of heat required = 1000 g × 334 J/g = 334000 joules
Step 4: Convert the heat from joules to kilocalories:
1 kilocalorie (kcal) is equivalent to 4184 joules.
Amount of heat in kilocalories = amount of heat (in joules) / 4184
Amount of heat in kilocalories ≈ 334000 J / 4184 ≈ 79.81 kcal
Rounding to the nearest whole number, the amount of heat removed by the ice as it melts to water at 0°C is approximately 80 kcal.
To calculate the amount of heat that the ice will remove as it melts to water, you can use the concept of heat transfer and the specific heat of fusion.
The specific heat of fusion, denoted as "L", is the amount of heat energy required to convert a substance from a solid to a liquid state without a change in temperature. For water, the specific heat of fusion is 80 calories per gram or 80 kcal/kg.
In your case, you have a 1 kg block of ice at 0 degrees Celsius, which means it is already at its melting point. So, to calculate the amount of heat that will be removed as the ice melts, you can multiply the mass of the ice (1 kg) by the specific heat of fusion (80 kcal/kg):
Heat removed = (mass of ice) x (specific heat of fusion)
= 1 kg x 80 kcal/kg
= 80 kcal
Therefore, the ice will remove 80 kcal of heat energy as it melts to water at 0 degrees Celsius.