what mass of water at 10.0 degrees C is needed to cool 500.0ml of tea from 25.0 degrees C to 18.3 degrees C? Density and specific heat of the tea are the same as water.
This is done the same way as adding cool water to warm water. The only difference is that you solve for mass cool water rather than Tfinal.
(? x 4.184 x 6.7) = (500 x 4.184 x 8.3)
28.03x = 17363.6
x= 6.19 x 10(2)g
is this correct?
I don't think so.
It appears to me that you've reversed the delta T values.
[? x 4.184 x (8.3)] = 500 x 4.184 x 6.7
mass 10 C H2O needed approximately 400 g if I didn't punch the wrong keys.
To solve this problem, we can use the equation for heat transfer:
q = mcΔT
Where:
q is the heat transferred (in joules)
m is the mass of the substance (in grams or kilograms)
c is the specific heat capacity of the substance (in J/g·°C or J/kg·°C)
ΔT is the change in temperature (in °C)
First, let's calculate the heat transferred when the tea is cooled:
q = mcΔT
We know the final temperature (18.3 degrees C) is lower than the initial temperature (25.0 degrees C), so ΔT = (18.3 - 25.0) = -6.7 degrees C.
Now, we need to determine the specific heat capacity (c) of water since it is the same as tea. The specific heat capacity of water is approximately 4.18 J/g·°C.
Next, we want to solve for the mass (m) of the water at 10.0 degrees C that is needed to cool the tea. To do this, we rearrange the equation:
m = q / (cΔT)
Substituting the given values:
m = (500.0 ml) x density of water x (10.0 - 18.3) / (4.18 J/g·°C)
Note that we need to convert the volume from milliliters (ml) to grams (g), since density is mass divided by volume.
To convert ml to g, we need to know the density of water. The density of water at 4 degrees Celsius is approximately 1 g/ml.
Density of water = 1 g/ml
So, multiply the volume (500.0 ml) by the density (1 g/ml) to get the mass in grams:
m = (500.0 ml) x (1 g/ml) x (10.0 - 18.3) / (4.18 J/g·°C)
Performing the calculation:
m = (500.0 g) x (-8.3) / (4.18 J/g·°C)
m = -996.4 g·°C / (4.18 J)
m ≈ -239 g
The negative sign indicates that the water lost heat while cooling the tea. In this case, we need to take the absolute value of the mass, so the mass of water needed to cool the tea from 25.0 degrees C to 18.3 degrees C is approximately 239 grams.