A 1.2 kg block of ice is initially at a temperature of -6.0 degrees C. A) If 2.2
ID: 1864161 • Letter: A
Question
A 1.2 kg block of ice is initially at a temperature of -6.0 degrees C. A) If 2.2x10^5 J of heat are added to the ice, what is the final temperature of the system?T=_____ degrees C
B) Find the amount of ice, if any, that remains.
m=______ kg A 1.2 kg block of ice is initially at a temperature of -6.0 degrees C. A) If 2.2x10^5 J of heat are added to the ice, what is the final temperature of the system?
T=_____ degrees C
B) Find the amount of ice, if any, that remains.
m=______ kg A) If 2.2x10^5 J of heat are added to the ice, what is the final temperature of the system?
T=_____ degrees C
B) Find the amount of ice, if any, that remains.
m=______ kg
Explanation / Answer
We have
Mass of block of ice,m = 1.2 Kg
Initial temperature of block = -6 degree Celsius
We know that
Specific heat capacity of water = 4.186 KJ/Kg.K
Latent heat of fusion = 334 KJ/Kg
Therefore amount of energy required to change temperature to zero degree Celsius = 4.186 * 1.2 * 6 = 30.14 KJ
A) amount of heat given to the block = 220 KJ
Therefore amount of heat that changes ice to water = 220 - 30.14 = 189.86 KJ
Therefore amount of ice converted water = heat given to ice at zero degree Celsius/latent heat of fusion = 189.86/334 = 0.5684 Kg
Therefore remaining amount of ice = 1.2 - 0.5684 = 0.6316 Kg
As ice is remained
Therefore final temperature of system will be zero degree Celsius
b) amount of ice remained = 0.6316 Kg......ANWER
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