An amt of ice was added to a 8.25 kg tubof water at 25.0°C. If their final temp is 18.2, what mass of ice was added?
Please help, I keep getting the wrong answer
To find the mass of ice that was added to the water, we can use the principle of conservation of energy. The energy lost by the water when it cools down to the final temperature must be gained by the ice when it melts.
To solve this problem, we can use the equation:
Qwater + Qice = 0
where Qwater represents the energy lost by the water and Qice represents the energy gained by the ice.
The energy lost by the water can be calculated using the formula:
Qwater = mcΔT
where m is the mass of water, c is the specific heat capacity of water, and ΔT is the change in temperature.
The energy gained by the ice can be calculated using the formula:
Qice = mLf
where m is the mass of ice and Lf is the latent heat of fusion.
Since we are trying to find the mass of ice, we can rearrange the equation to solve for m:
m = -Qwater / Lf
Now let's calculate the values needed to solve for the mass of ice:
Given:
Mass of water (mwater) = 8.25 kg
Initial temperature of water (Tinitial) = 25.0°C
Final temperature (Tfinal) = 18.2°C
Specific heat capacity of water (cwater) = 4.186 J/g°C (this needs to be converted to J/kg°C)
Latent heat of fusion (Lf) = 3.34 x 10^5 J/kg
First, let's convert the specific heat capacity of water from J/g°C to J/kg°C:
1 g = 0.001 kg (conversion factor)
cwater = 4.186 J/g°C × 0.001 kg/g
cwater = 0.004186 J/kg°C
Now, let's calculate the energy lost by the water (Qwater):
Qwater = mwater × cwater × ΔT
ΔT = Tfinal - Tinitial
ΔT = 18.2°C - 25.0°C
ΔT = -6.8°C (Note: We use a negative sign because the temperature is decreasing)
Qwater = 8.25 kg × 0.004186 J/kg°C × -6.8°C
Once you calculate Qwater, substitute the value into the rearranged equation to solve for the mass of ice (m):
m = -Qwater / Lf
Now, solve for m using the given values. The resulting value will be the mass of ice that was added to the tub of water.