When 80g of water of 40° C is mixed with 40g ice of 0°C, the common temperature is ..
1) 0°C
2) 4°C
3) 6°C
4) 12°C
Select one:
1
ОО
4
3
2
2
d
Answers
Answer:
4. is the common temprature ...
Understanding the question -
=> Here, 80 g of water at 40 °C is mixed with 40 g of ice at 0 °C .
=> Heat transfer takes place from the water to the ice until the same temperature is attained.
=> We need to find the final temperature of the mixture assuming that no heat is lost to the surroundings or vice-versa .
Values that you should know -
- Latent heat of ice is 336,000 J kg⁻¹ .
=> This means that in converting 1 kilogram of ice at 0° C to 1 kilogram of water at 0 °C, 336,000 joules of heat is absorbed.
- Specific heat capacity of water is 4200 J kg⁻¹ °C⁻¹ .
=> This means that in lowering the temperature of 1 kilogram of water by 1 K (or 1 °C) 4200 joules of heat energy is released.
Concepts/Formulae Used Here -
=> The concept of heat transfer is used here -
- Q = m . c . △T
Where -
Q = amount of heat lost/gained
m = mass of substance
c = specific heat capacity
△T = Change in temperature
=> The concept of latent heat of melting is also used -
Where -
L = Latent heat of melting (per unit mass)
Q = Amount of heat absorbed
m = Mass of substance
The steps we will use and solving accordingly -
1) First, we determine the amount of heat that the ice at 0 °C will need to melt completely to give water at 0 °C -
=> For this, we use the formula .
L = 336,000 J kg⁻¹
Q = ?
m = 40 g
=>
=>
=>
=> 336 * 40 = Q
=> Q = 13,440 joules
=> Thus, the amount of heat required to melt the ice completely is 13,440 J .
=> We know that all of this heat (13,440 J) will have to be supplied by the water.
=> The ice gains 13,440 J of heat and changes into water at 0 °C .
2) Now, we find what temperature the water will reach after supplying 13,440 joules to the ice -
=> For this we use the formula - Q = m . c . △T
Q = 13,440 J
m = 80 grams
c = 4200 J kg⁻¹ °C⁻¹
△T = T1 - T2
T1 = 40 °C (initial temperature)
T2 = ? (final temperature)
Let T2 be x .
△ T = (40-x)
=> 13,440 = 0.08 * 4200 * (40-x)
=> 13,440 = 336(40-x)
=> 40-x = 13,440/336
=> 40-x = 40
=> 40-40 = x
=> x = 0 °C
=> T2 = 0 °C
=> Thus, the final temperature of the water after losing 13,440 J will be 0 °C .
3) Now we have -
‣ 40 grams of water (formed from the ice) at 0 °C .
‣ 80 grams of water (formed after losing 13,440 J to ice) at 0 °C .
=> Since both are now at the same temperature no heat transfer takes place.
Answer -
=> The final temperature of the mixture will be water at 0 °C .
=> Thus, the correct answer is option (1) 0°C .
More to know -
=> The specific heat capacity of ice is about 2100 J kg⁻¹ °C⁻¹ .
- This means that in increasing/decreasing the temperature of one kilogram sample of ice by 1 °C or 1 K the amount of heat required is 2100 joules .
=> The specific heat capacity of steam is about 2000 J kg⁻¹ °C⁻¹ .
- This means that in increasing/decreasing the temperature of one kilogram sample of steam by 1 °C or 1 K the amount of heat required is 2000 joules .