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
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6694.4 J = (20 grams H2O)(4.180 J/gC)(Tf - 0o C)
6694.4/83.6 = Tf
= 80o Celsius
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The final temperature would be approximately 54.2 degrees Celsius. This can be calculated using the principle of conservation of energy, where the heat lost by the hot water is equal to the heat gained by the cold water.
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.
The amount of water in fog over a city block can vary widely depending on factors like temperature, humidity, and fog density. However, on average, fog can contain around 0.05 to 0.5 grams of water per cubic meter. This means that over a city block, which can cover an area of 10,000 to 20,000 square meters, there could be between 500 to 10,000 grams (or 0.5 to 10 kilograms) of water in the fog.
Grams are mass. There's no calculation involved.
34.7 ml of O2 at temperature 0 deg Celsius and 101kPa pressure and contains 0.0496 grams or 49.6 milligrams of oxygen.This can be worked out by multiplying 34.7 ml by oxygen's density of 0.001429 g/ml.
The final temperature is 59.9°C.
The volume depends on the temperature and pressure - neither of which are given.
The maximum amount of solute that can be dissolved in a solvent at a given temperature is known as the solubility of that solute in the solvent. Solubility is typically expressed in terms of concentration, such as grams of solute per 100 grams of solvent. It varies with temperature; generally, solubility increases with temperature for solids, while it may decrease for gases.
To calculate the final temperature, you need to use the formula: q = mcΔT, where q is the heat energy, m is the mass, c is the specific heat capacity of water, and ΔT is the change in temperature. Rearrange the formula to solve for the final temperature Tf: Tf = (q / (m*c)) + Ti, where Ti is the initial temperature. Plug in the values and calculate the final temperature.
The final temperature would be approximately 54.2 degrees Celsius. This can be calculated using the principle of conservation of energy, where the heat lost by the hot water is equal to the heat gained by the cold water.
It is not possible to answer the question in which the volume of the block is given as Cm Cm and Cm. Some numbers might have been useful!
As the volume of a given gas sample is dependent on its temperature and pressure; to find a volume of any gas which does exist, the temperature and the pressure of the system/vessel should be given directly or could be calculated.
About 232 grams
I'll assume here that by "70 temperature" you mean "70 degrees Celsius". Basically, you have to calculate the average temperature of all of the water in the mixture, which will be the final temperature once it's well stirred. The 200 grams of water at 10 degrees represent 2/3 of the total amount of water (300 grams), so thus, multiply 10 by 2/3 to determine their contribution to the final temperature. You will get 20/3. The 100 grams of water at 70 degrees represent 1/3 of the total amount of water, so multiply 70 by 1/3 to determine their contribution to the final temperature. You will get 70/3. When you add together the two temperatures you get 90/3, which is equal to 30. Therefore, the final temperature is 30 degrees Celsius.
A standard building block typically weighs around 200-300 grams depending on its size and material.
To find the temperature in kelvins required to dissolve 400 grams of sugar in 100 grams of water, you would typically refer to solubility data for sugar in water. This data indicates how much sugar can dissolve at various temperatures. Since solubility increases with temperature, you would need to heat the water until the desired amount of sugar fully dissolves. The final temperature in kelvins can be calculated by adding 273.15 to the Celsius temperature obtained from the solubility data.
Without additional information given such as temperature, it is difficult to determine the solubility of 618 grams of zinc chloride in 150 cm3 of water as the solubility of zinc chloride in water varies with temperature. Typically, at room temperature, around 432 grams of zinc chloride can dissolve in 150 cm3 of water.