In prepairing a tea 600g of water at 90 degrees celsius is poured into a 200g China pot.Cpot=840 J/kg degrees celsius at 20 degrees celsius. What is the final temperature of water?
Need answer
To find the final temperature of the water, we can use the principle of conservation of energy. The heat gained by the water will be equal to the heat lost by the China pot.
The heat gained by the water can be calculated using the formula:
Qwater = mwater * cwater * (Tf - Ti)
where:
Qwater = heat gained by the water
mwater = mass of water (600g)
cwater = specific heat capacity of water (4.18 J/g°C)
Tf = final temperature of the water
Ti = initial temperature of the water (90°C)
The heat lost by the China pot can be calculated using the formula:
Qpot = mpot * cpot * (Tf - Ti)
where:
Qpot = heat lost by the China pot
mpot = mass of the China pot (200g)
cpot = specific heat capacity of the China pot (840 J/kg°C)
Tf = final temperature of the water
Ti = initial temperature of the water (90°C)
Since the heat gained by the water is equal to the heat lost by the China pot, we can set up an equation:
mwater * cwater * (Tf - Ti) = mpot * cpot * (Tf - Ti)
Plugging in the values:
600g * 4.18 J/g°C * (Tf - 90°C) = 200g * 840 J/kg°C * (Tf - 90°C)
Simplifying the equation:
2508(Tf - 90) = 336000(Tf - 90)
Dividing both sides by (Tf - 90):
2508 = 336000
Now we can solve for Tf:
2508 / 336000 = Tf - 90
0.0075 = Tf - 90
Adding 90 to both sides:
Tf = 90 + 0.0075
Tf ≈ 90.0075°C
Therefore, the final temperature of the water is approximately 90.0075°C.
To find the final temperature of the water, we can use the principle of conservation of energy:
The amount of heat lost by the water = the amount of heat gained by the China pot.
Heat lost by the water can be calculated using the formula:
Q = m * c * ΔT
Where:
Q = heat lost by the water
m = mass of the water (600g)
c = specific heat capacity of water (4.18 J/g°C)
ΔT = change in temperature of the water (final temperature - initial temperature)
Heat gained by the China pot can be calculated using the formula:
Q = m * c * ΔT
Where:
Q = heat gained by the China pot
m = mass of the China pot (200g)
c = specific heat capacity of the China pot (840 J/kg°C)
ΔT = change in temperature of the China pot (final temperature - initial temperature)
Since the water and the China pot are in thermal equilibrium at the final temperature, we can equate the two equations:
m_water * c_water * ΔT_water = m_pot * c_pot * ΔT_pot
Substituting the given values:
(600g) * (4.18 J/g°C) * (ΔT_water) = (200g) * (840 J/kg°C) * (ΔT_pot)
Now we can solve for ΔT_water, which represents the change in temperature of the water:
(ΔT_water) = [(200g) * (840 J/kg°C) * (ΔT_pot)] / [(600g) * (4.18 J/g°C)]
(ΔT_water) = (112,000J * ΔT_pot) / (2,508J)
Simplifying:
(ΔT_water) ≈ 44.62 ΔT_pot
Since the water was initially at 90°C, we can write the equation:
(ΔT_water) + (90°C) = (ΔT_water) + (ΔT_pot) + (20°C)
Simplifying:
90°C = (ΔT_water) + (ΔT_pot) + 20°C
Rearranging the equation:
(ΔT_water) + (ΔT_pot) = 90°C - 20°C
Substituting the value of ΔT_water from the previous equation:
44.62 ΔT_pot + (ΔT_pot) = 70°C
Simplifying:
45.62 ΔT_pot = 70°C
ΔT_pot ≈ 1.5373°C
Now we can find the final temperature of the water:
Final temperature = initial temperature + ΔT
Final temperature = 90°C + 1.5373°C
Final temperature ≈ 91.54°C
Therefore, the final temperature of the water is approximately 91.54°C.