I believe it will be 145.52 degrees Celsius if I did my math correctly. You need to convert calories to joules. I believe one joule raises the temp of 1 gram water by 1 degree Celsius so 1200*4.184=5020.8 J /40grams=125.52 temp increase+20=145.52 degrees Celsius.
40 degree
Density = (Pressure)(Molar Mass)/(R)(Temperature) I forget exactly what R is called, but it is a constant. So, you get (.970)(46)/(.0821)(308). Your final answer is 1.76456. Round to as many decimal places as your teacher prefers
4.0%
I must assume that the ice block is at an initial temperature of 0o C. I also need to work in joules,. so I will convert the calories to joules. 1600 cal. (4.184 joule/1 cal.) = 6694.4 joules -------------------- 6694.4 J = (20 grams H2O)(4.180 J/gC)(Tf - 0o C) 6694.4/83.6 = Tf = 80o Celsius ==========
-5.58 C
1600
This depends on the volume.
No, as both the temperatures are the same, you will get only 2 cups, each 50 degrees. You have to heat the cup to get 100 degree.
The metal with the lowest thermal capacity.
42.3 C
Do some converting first. 688 calories (4.184 Joules/1 calorie) = 2878.592 Joules 25 ml of water = 25 grams q(Jolules) = mass * specific heat * (Temp. final - Temp. initial) 2878.592 Joules = 25 grams Water * 4.180 J/gC * (Temp Final - 80C ) 2878.592 Joiles = 104.5( Temp. Final) - 8360 11238.592 =104.5(Temp. Final) 107.55 Celsius Final Temperature ( call it 108 C )
When hot metal is added into the water then the metal looses its energy into the water and this heat is gained by the water, so the temperature gets increases when hot metal added into it i.e final temperature is greater than initial temperature of water.
I need to make some conversions for my own convenience. 750 calories (4.184 Joules/1 calorie) = 3138 Joules ----------------------- Lead specific heat (c) = 0.160 J/gC Now, q = mass * specific heat * change in temperature 3138 Joules = (250 g Pb)(0.160 J/gC)(Tf - 28.0 C) 3138 J = 40Tf- 1120 4258 = 40Tf 106 Celsius = Final temperature of lead ----------------------------------------------------
21
This is the formula for finding calories: calories= (Tf-Ti)mass Or, in words, you subtract the final temperature (the temperature of the fused waters) by its initial temperature (the hot/ cold water's original temperature). You then multiply this answer by the mass of the water before you mix them together. Your hot and cold water should both have the same mass, but not the same temperature.
if 2.5kg of hot water at 100c is added to 10kg of cold water at 28c and stirred well. what is the final temperature of mixture? (neglect the heat absorbed by container and the heat lost by the surroundings.)
980000/ (6200*4.180)+18`c or 291K final answer should be 56`c or 329K.