150g of ice at 0'c is mixed with 300g of water at 50'c. calculate the temperature of a mixture
To calculate the temperature of the mixture, you can use the principle of conservation of energy. The energy lost by the hot water will be equal to the energy gained by the ice.
To start calculations, we need to determine the heat lost by the hot water and the heat gained by the ice.
The heat lost by the hot water can be calculated using the formula:
Q(hot water) = mass(hot water) * specific heat capacity(water) * change in temperature
Where:
mass(hot water) = 300g
specific heat capacity(water) = 4.186 J/g°C (specific heat capacity of water)
change in temperature = final temperature - initial temperature = TF - 50°C
The heat gained by the ice can be calculated using the formula:
Q(ice) = mass(ice) * specific heat capacity(ice) * change in temperature
Where:
mass(ice) = 150g
specific heat capacity(ice) = 2.09 J/g°C (specific heat capacity of ice)
change in temperature = final temperature - initial temperature = TF - 0°C
Since the heat lost by the hot water is equal to the heat gained by the ice, we can set up an equation:
mass(hot water) * specific heat capacity(water) * change in temperature(hot water) = mass(ice) * specific heat capacity(ice) * change in temperature(ice)
300g * 4.186 J/g°C * (TF - 50°C) = 150g * 2.09 J/g°C * (TF - 0°C)
Now, we can solve this equation to find the final temperature (TF):
(300g * 4.186 J/g°C * TF) - (300g * 4.186 J/g°C * 50°C) = (150g * 2.09 J/g°C * TF) - (150g * 2.09 J/g°C * 0°C)
300g * 4.186 J/g°C * TF - 300g * 4.186 J/g°C * 50°C = 150g * 2.09 J/g°C * TF - 150g * 2.09 J/g°C * 0°C
1255.8g°C * TF - 62790g°C = 313.5g°C * TF
(1255.8g°C - 313.5g°C) * TF = 62790g°C
942.3g°C * TF = 62790g°C
TF = 62790g°C / 942.3g°C
TF = 66.6°C
Therefore, the temperature of the mixture will be approximately 66.6°C.
To calculate the temperature of a mixture, we can use the principle of conservation of energy. The formula we'll use is:
(m1 * c1 * ΔT1) + (m2 * c2 * ΔT2) = 0
Where:
m1 is the mass of the first substance (ice)
c1 is the specific heat capacity of the first substance (ice)
ΔT1 is the change in temperature of the first substance (final temperature - initial temperature)
m2 is the mass of the second substance (water)
c2 is the specific heat capacity of the second substance (water)
ΔT2 is the change in temperature of the second substance (final temperature - initial temperature)
First, let's find the change in temperature for each substance:
For the ice:
Mass (m1) = 150g
Specific heat capacity (c1) = 2.09 J/g°C (specific heat capacity of ice)
Initial temperature = 0°C
Final temperature = ?
For the water:
Mass (m2) = 300g
Specific heat capacity (c2) = 4.18 J/g°C (specific heat capacity of water)
Initial temperature = 50°C
Final temperature = ?
Now, plug the values into the formula:
(150g * 2.09 J/g°C * ΔT1) + (300g * 4.18 J/g°C * ΔT2) = 0
Simplifying the equation:
313.5 ΔT1 + 1254 ΔT2 = 0
Since the ice and water are mixed, the final temperature of the mixture will be the same. Therefore, ΔT1 = ΔT2
313.5 ΔT + 1254 ΔT = 0
1567.5 ΔT = 0
ΔT = 0
Since ΔT = 0, it means there is no change in temperature during the mixing process. Therefore, the temperature of the mixture will be the initial temperature of the water, which is 50°C.
sum heats transfered is zero (some are + gains, some are - gains).
heat to melt ice+heat to heat melted ice + heat to original water=0
Lf=latent heat of fusion ice
cw=specific heat capacity of water
Lf*150+150*cw(Tf-0)+300*cw*(Tf-50)=0
look up Lf
solve for Tf