Suppose 100 g of water at 20°C is poured over a 70-g cube of ice with a temperature of -8°C. If all the ice melts, what is the final temperature of the water? If all of the ice does not melt, how much ice remains when the water–ice mixture reaches equilibrium?
I'm aware that not of the ice melts, but how do I determine how much ice is left?
You balance the heat equation.
heat gained by melting ice+heat gained by remaining ice warming up + heat gained by hot water=0
This of course means that the nonmelted ice is at a final temperature of zero, otherwise there cannot be an ice/water mxture.
Heat gained by melting ice:
(70-m)Hf+ (70-m)cice(8)
Heat gained by nonmelting ice:
(m*cice(8)
heat gained by hot water: 100*cwater(0-20)
add them, set to zero, solve for m.
I'm getting -25.1 g when I use this equation.
.07*334kJ-m334+.07*2.1*8+.1*4.18*(-20)=0
m= - 1/334 * (.418*20-.07*334-.07*16.8)
which is not a negative number....
check my work, most of it was in my head.
I forgot that 70g was 0.07 kg. I was using 0.007 kg. Thank you. This question makes a lot more sense now.
7.896 degrees celsius
7.896 degrees celsius and 15.125 g of ice
To determine how much ice is left when the water-ice mixture reaches equilibrium, you can use the concept of heat transfer.
First, let's calculate the amount of heat gained by the ice to reach its melting point. This can be done using the equation:
Q = m * c * ΔT
Where:
Q : heat gained or lost (in Joules)
m : mass of the substance (in grams)
c : specific heat capacity of the substance (in J/g°C)
ΔT : change in temperature (in °C)
For the ice, the change in temperature is the difference between its initial temperature (-8°C) and its melting point (0°C), which is 8°C. The specific heat capacity of ice is 2.09 J/g°C.
Q_ice = (70 g) * (2.09 J/g°C) * (8°C)
Q_ice = 1171.6 J
Next, we need to calculate the amount of heat required to melt the ice. This can be done using the equation:
Q = m * ΔH_fusion
Where:
Q : heat gained or lost (in Joules)
m : mass of the substance (in grams)
ΔH_fusion : heat of fusion (melting) of the substance (in J/g)
For ice, the heat of fusion is 333.5 J/g.
Q_melt = (70 g) * (333.5 J/g)
Q_melt = 23345 J
To find the total heat gained by the ice, we sum up the heat gained to reach melting point (Q_ice) and the heat required to melt the ice (Q_melt).
Total heat gained by ice = Q_ice + Q_melt
Total heat gained by ice = 1171.6 J + 23345 J
Total heat gained by ice = 24477.6 J
Now, assuming that all the ice does not melt, we can use the concept of heat transfer to find the final temperature of the water. The heat lost by the water is equal to the total heat gained by the ice.
Q_lost = Total heat gained by ice = 24477.6 J
Using the equation Q = m * c * ΔT, where ΔT is the change in temperature for the water, c is the specific heat capacity of water (4.18 J/g°C), and m is the mass of the water (100 g), we can solve for ΔT.
24477.6 J = (100 g) * (4.18 J/g°C) * ΔT
Dividing both sides by (100 g) * (4.18 J/g°C):
ΔT = (24477.6 J) / [(100 g) * (4.18 J/g°C)]
ΔT = 58.7°C
Therefore, if all of the ice does not melt, the final temperature of the water will be 58.7°C.