If 100 ml of water at 60 degree is added to 200 ml of water at 30 degree then the temperature in mixture is options A) 45 degree B) 40 degree C) 50 degree
Temp in mixture =M1t1+m2t2/m1+m2
100*60+200*30/100+200=
6000+6000 /300=
12000/300=
40
Hence,40 is the temperature in mixture.
Hope this helps you.
To find the temperature of the mixture, we can use the principle of conservation of energy.
Step 1: Calculate the heat absorbed by the hot water:
The formula to calculate heat is Q = mcΔT, where Q is the heat absorbed, m is the mass of the substance, c is its specific heat capacity, and ΔT is the change in temperature.
Given:
Mass of water A (100 ml) = 100 g (since 1 ml of water is approximately 1 g)
Temperature of water A = 60°C
Specific heat capacity of water = 4.18 J/g°C (approximate value)
Using the formula, we can find the heat absorbed by water A:
Q(A) = m(A) * c * ΔT(A)
Q(A) = 100 g * 4.18 J/g°C * (60°C - Tf)
Q(A) = 418 J/g°C * (60 - Tf)
Step 2: Calculate the heat gained by the cold water:
Given:
Mass of water B (200 ml) = 200 g
Temperature of water B = 30°C
Using the same formula, we can find the heat gained by water B:
Q(B) = m(B) * c * ΔT(B)
Q(B) = 200 g * 4.18 J/g°C * (Tf - 30°C)
Q(B) = 836 J/g°C * (Tf - 30)
Step 3: Since energy is conserved, the heat absorbed by A is equal to the heat gained by B. Therefore:
Q(A) = Q(B)
418 J/g°C * (60 - Tf) = 836 J/g°C * (Tf - 30)
Simplifying the equation:
(60 - Tf) = 2 * (Tf - 30)
60 - Tf = 2Tf - 60
3Tf = 120
Tf = 40
Therefore, the temperature of the mixture is 40°C (option B).
To find the temperature of the mixture, we need to use the principle of heat transfer, which states that the heat lost by one substance is equal to the heat gained by another substance.
Let's calculate the heat gained by the cooler water and the heat lost by the hotter water using the formula:
Q = mcΔT
Where:
Q is the amount of heat gained or lost
m is the mass of the substance
c is the specific heat capacity of the substance
ΔT is the change in temperature
For the cooler water:
m = 200 ml = 200 g (since the density of water is 1 g/ml)
c = 1 cal/g°C (specific heat capacity of water)
ΔT = final temperature (Tf) - initial temperature (Ti) = Tf - 30°C
Q1 = 200 × 1 × (Tf - 30)
For the hotter water:
m = 100 ml = 100 g
c = 1 cal/g°C
ΔT = final temperature (Tf) - initial temperature (Ti) = Tf - 60°C
Q2 = 100 × 1 × (Tf - 60)
Since heat gained = heat lost, we can equate the two equations:
Q1 = Q2
200 × 1 × (Tf - 30) = 100 × 1 × (Tf - 60)
Simplifying the equation:
200Tf - 6000 = 100Tf - 6000
100Tf = 200Tf - 6000
Combining like terms:
100Tf = 200Tf - 6000
-100Tf = -6000
Tf = 60°C
Therefore, the temperature of the mixture is 60 degrees Celsius.
Heat lost = Heat gained
Let x be the final temperature of the mixture.
(100)(60 - x)(4.184)=(200)(x - 30)(4.184)
Solve for x