help on physics homework please?

Question

Ice at 0.0°C is mixed with 5 × 10^2 mL of water at 28°C. How much ice must melt to lower the water temperature to 0.0°C?

i've done this kind of problem before given the amount of water in g, but how do you do it when given in mL??

thanks!!

thanks you for telling me that 1 mL of water equal 1 g... with that i figured out that I would need 175.7 g of ice to melt!! thanks :)

Answer 1

Question 2 first, unless you need 3 place accuracy, 1 g of water = 1 mL. Here, you have to transfer w(h2o) x Cp x temp change calories to melt ice. So the calories available are 500x28= 14000 calories. If I remember, 1 gram of ice needs 144 calories to melt at 0 degC. So you would need about 95 g of ice.

Answer 2

Ice; 0C/ ? ml : H2O; 28C/5 * 10^2ml.



(5 * 10^2ml)/28C = 17.857ml of water = 1C . Or 17.857 * 28 = 500ml.

So, it takes 1C to raise 17.857ml of waters temperature by 1C. Since there are 500ml of water at 28C. There must also be 500ml of ice at 0C to lower the 500 ml of water at 28C down to 0C.
The mass and volume of water has to be the same in this case,since water and ice have the same Chemical properties.
Ask yourself: If I weigh 200lbs and I'm on a teeter tater with my friend and I want for both of us to be (leveled) at 90 degrees or even on it. One must not be either lower or higher than the other one. Then, how much must my friend weigh ?

Answer 3

The density of water is 1000 kg/ m^3 (cubic meter) therefore just do simple conversion: If 1L = 0.0001m^3
1000L = 1m^3
density of water is 1000 kg/1000L = 1kg/1L. We have 5 x10^2 mL = 0.5L, therefore 1 kg/L * 0.5L = 0.5Kg = 500grams
Thus mL = grams, for water only due to the liquid density of the substance.

Now you can solve the problem the way you know.