How many grams of ice must be added to 100 grams of water at 60 degrees so that its temperature drops to 20 degrees celcius?
As far as i know I have to use:
(Mass)*(specific heat of water) *(finaltemperature - initial temperature = ??
What is the temperature of the ice when you put it in?
heat to lose=100*specifheatwater*40
heat absorbed by ice: mass*specificheatice*(Ti)+ mass*Heatfusionice*(20)
set them equal, after putting in Tinitial of ice, and solve for the mass of ice.
the temperature of ice according to our physics teacher is -5 degrees celcius
Then
massice*(specificheatice)5+massice*heatfusionice*20= 100*spcificheatdwater*40
To calculate the amount of ice that needs to be added, we can use the equation:
(mass of water) * (specific heat of water) * (final temperature - initial temperature) = (mass of ice) * (specific heat of ice) * (final temperature - initial temperature)
In this case, the initial temperature of the water is 60 degrees Celsius, and the final temperature needed is 20 degrees Celsius. The specific heat of water is 4.18 J/g°C, and the specific heat of ice is 2.09 J/g°C.
Let's plug in the values we know:
(100 g) * (4.18 J/g°C) * (20°C - 60°C) = (mass of ice) * (2.09 J/g°C) * (20°C - 0°C)
Simplifying the equation:
-2080 J = (mass of ice) * (2.09 J/g°C) * 20°C
Now, let's solve for the mass of ice:
mass of ice = -2080 J / [(2.09 J/g°C) * 20°C]
mass of ice = -2080 J / (41.8 J/g)
mass of ice = -49.76 g
Notice that the result is negative, indicating that the ice needs to be removed from the system rather than added. This means that the 100 grams of water at 60 degrees Celsius will naturally cool down to 20 degrees Celsius without adding any ice.