q = m x C x DT
q = amount of heat energy gained or lost in Joules (J)
m = mass of sample in grams = 50g
C = heat capacity (J/ g•oC) = 4.186J/g•oC for water
Tf = final temperature = 95 oC
Ti = initial temperature = 40 oC
DT = (Tf - Ti) = (95 oC - 40 oC) = 55 oC
q = 50g x 4.186J/g•oC x 55 oC = 11511.5 J = 10000 J (rounded to 1 significant figure)
The amount of solute that will dissolve in water at a specific temperature depends on the solubility of the solute at that temperature. You need to check the solubility of the specific solute at the specified temperature (in degrees Celsius) to determine how many grams will dissolve in 100 grams of water.
The amount of salt that can be dissolved in 200ml of water at room temperature depends on the solubility of salt in water. Typically, at room temperature, you can dissolve about 36 grams of salt in 200ml of water.
Sodium nitrate is soluble in water at any temperature. It has high solubility, and 120 grams of sodium nitrate will dissolve in 100 ml of water regardless of the temperature.
Approximately 180 grams of sodium nitrate can be dissolved in 100 grams of water at 50°C.
To calculate the number of joules of heat gained by water, you can use the formula Q = m * c * ΔT, where Q is the heat gained, m is the mass of water in grams, c is the specific heat capacity of water (4.18 J/g°C), and ΔT is the change in temperature in degrees Celsius. Plug in the values for m, c, and ΔT to calculate the heat gained in joules.
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.
The answer depends on the temperature, but at room temperature (20 deg C), 100 ml of water would have a mass of 99.82 grams.
Heat is lost to or gained from the air while falling.
The amount of solute that will dissolve in water at a specific temperature depends on the solubility of the solute at that temperature. You need to check the solubility of the specific solute at the specified temperature (in degrees Celsius) to determine how many grams will dissolve in 100 grams of water.
To solve this problem, you can use the principle of conservation of energy: Heat gained by the metal = heat lost by the metal + heat gained by the water Plug in the given values and use the specific heat capacity of metal and water to calculate the initial temperature of the metal before mixing.
air from the lake
The amount of salt that can be dissolved in 200ml of water at room temperature depends on the solubility of salt in water. Typically, at room temperature, you can dissolve about 36 grams of salt in 200ml of water.
The energy needed to change ice into water is called the heat of fusion. For ice, this value is around 334 joules per gram. So, for 3 grams of ice, the energy gained when it changes to water would be around 1002 joules (334 joules/gram * 3 grams).
Sodium nitrate is soluble in water at any temperature. It has high solubility, and 120 grams of sodium nitrate will dissolve in 100 ml of water regardless of the temperature.
To calculate the number of joules of heat gained by water, you can use the formula Q = m * c * ΔT, where Q is the heat gained, m is the mass of water in grams, c is the specific heat capacity of water (4.18 J/g°C), and ΔT is the change in temperature in degrees Celsius. Plug in the values for m, c, and ΔT to calculate the heat gained in joules.
Approximately 180 grams of sodium nitrate can be dissolved in 100 grams of water at 50°C.
For fresh water at room temperature, it is 25 grams