How much heat is absorbed by 100 g of ice at -20 deg C to become water at 0 deg C? The specific heat of ice is 2.06 J/gC.
q1 to change T ice @ -20 C to ice @ zero C.
q1 = mass ice x specific heat ice x (Tfinal-Tinitial) = ?
q2 = heat to change ice @ zero C to water @ zero C.
q2 = mass ice x heat fusion ice = ?
total Q = q1 + q2 = ?
Well, let's take a moment to appreciate the irony here. We're talking about something going from cold to colder, which is pretty unique. Now, to get back to the question, we know that to warm up ice, we need to add heat. The specific heat of ice is 2.06 J/g°C, which means for every gram of ice, we need to add 2.06 joules of heat for each degree Celsius increase. So, if we have 100 grams of ice and want to raise its temperature by 20 degrees Celsius, we'll need to do a little math. 100 grams times 2.06 joules per gram per degree Celsius times 20 degrees Celsius... and voila! We get 4120 joules of heat absorbed. So, that's how much heat you'd need to go from -20°C to 0°C. Now, I hope that explanation didn't give you the chills!
To calculate the heat absorbed by the ice, you need to consider the following steps:
Step 1: Calculate the temperature difference
The temperature difference is the change in temperature from -20°C to 0°C.
ΔT = final temperature - initial temperature
ΔT = 0°C - (-20°C)
ΔT = 20°C
Step 2: Calculate the heat absorbed
The heat absorbed is calculated using the formula:
Q = m * c * ΔT
where:
Q = heat absorbed (in Joules)
m = mass of the ice (in grams)
c = specific heat capacity of ice (in J/g°C)
ΔT = temperature difference (in °C)
Given:
m = 100 g
c = 2.06 J/g°C
ΔT = 20°C
Plugging the values into the formula:
Q = (100 g) * (2.06 J/g°C) * (20°C)
Q = 4120 J
Therefore, 100 g of ice at -20°C absorbs 4120 Joules of heat to become water at 0°C.
To calculate how much heat is absorbed by 100 g of ice at -20°C to become water at 0°C, you can use the formula:
Heat absorbed = mass × specific heat × change in temperature
In this case, the mass of the ice is 100 g, the specific heat of ice is 2.06 J/g°C, and the change in temperature is 0°C - (-20°C) = 20°C.
Plugging in these values into the formula:
Heat absorbed = 100 g × 2.06 J/g°C × 20°C
To get the answer, simply multiply the values together:
Heat absorbed = 4120 J
Therefore, 100 g of ice absorbs 4120 joules of heat to become water at 0°C.