How much heat must be removed from 5.0g of water at 10°C to become ice at 0°C ?
10°c to 0°c
Ht=mLf+mc(Tf-Ti)
m=5.0g
Lf=80cal/g
c=1cal/g°C
Ht=5.0(80)+(5.0)(1)(0-10)
Ht=400-50
=350cal
Well, that's a cool question! To answer it, we need to know the specific heat capacity of water and the heat of fusion for water. But don't worry, I'm here to melt those numbers away...with humor, of course!
So, let's break the ice! The specific heat capacity of water is 4.18 J/g°C. That means it takes 4.18 joules of heat energy to raise the temperature of 1 gram of water by 1 degree Celsius.
Now, the heat of fusion for water is 334 J/g, which is the amount of heat energy needed to convert 1 gram of water at 0°C into ice at 0°C.
So, first, let's raise the temperature of the water from 10°C to 0°C. That requires cooling it down by 10°C.
Heat removed = mass of water × specific heat capacity × change in temperature
= 5.0g × 4.18 J/g°C × 10°C
= 209 Joules
Now that we have reached 0°C, we need to convert the water into ice.
Heat removed = mass of water × heat of fusion
= 5.0g × 334 J/g
= 1670 Joules
Now, let's combine the two:
Total heat removed = heat to lower temperature + heat of fusion
= 209 Joules + 1670 Joules
= 1879 Joules
So, to steal the heat away from our 5.0g of water and turn it into ice, we need to remove 1879 Joules of heat. There you have it, a chilled answer for a cool question!
To find out how much heat must be removed from 5.0g of water at 10°C to become ice at 0°C, we can use the formula:
Q = m * C * ΔT
where Q is the amount of heat, m is the mass of the substance, C is the specific heat capacity, and ΔT is the change in temperature.
First, let's determine the heat required to cool the water from 10°C to 0°C. The specific heat capacity of water is approximately 4.18 J/g°C, which means that it takes 4.18 Joules of heat to raise the temperature of 1 gram of water by 1 degree Celsius.
ΔT = (Final temperature - Initial temperature) = (0°C - 10°C) = -10°C
Q = m * C * ΔT
= 5.0g * 4.18 J/g°C * (-10°C)
= -209 Joules (since we are removing heat, the value is negative)
Therefore, the amount of heat that must be removed from 5.0g of water at 10°C to become ice at 0°C is approximately 209 Joules.