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How much heat is necessary to change 565 g of ice at -13 and degC to water at 20 and degC?
To calculate the heat needed to change ice to water, we use the specific heat capacities and latent heat of fusion. First, calculate the heat needed to raise the temperature of 565 g of ice from -13°C to 0°C using specific heat capacity of ice. Then, calculate the heat needed to melt the ice at 0°C to water at 0°C using the latent heat of fusion for ice. Finally, calculate the heat needed to raise the temperature of water from 0°C to 20°C using the specific heat capacity of water. Add these three values together to find the total heat required.
What else can I help you with?
How much heat in joules would you need to raise the temperature of 1kg of water by 5 and degC?
You would need 20,920 Joules of heat to raise the temperature of 1kg of water by 5°C. This value is calculated using the specific heat capacity of water, which is 4186 J/kg°C.
What is the change in thermal energy of the air when it warms from 24 and degC to 31 and degC?
The change in thermal energy can be calculated using the formula: ( Q = mc\Delta T ), where ( Q ) is the change in thermal energy, ( m ) is the mass of the air, ( c ) is the specific heat capacity of air, and ( \Delta T ) is the change in temperature. The specific heat capacity of air is approximately 1000 J/kg°C. Assuming the mass is 1 kg, the change in thermal energy would be ( Q = 1kg \times 1000 J/kg°C \times 7°C = 7000 J ).
What does latent heat causes?
Latent heat has the ability to do something in a given time period. Take a water heated radiator, latent heat has the ability to make the radiator warm or hot according to the temprature of the water. It will continue to do this until the water stops flowing through the radiator and the radiator begins to cool.
What change of state happens when heat is removed from water and it condenses?
When heat is removed from water, it undergoes a change of state from a gas (water vapor) to a liquid (water). This process is known as condensation.
How are the amount of heat transfered and the change in temperature of water related?
The amount of heat transferred to water is directly proportional to the change in temperature it undergoes. This relationship is given by the equation Q = mcΔT, where Q is the amount of heat transferred, m is the mass of water, c is the specific heat capacity of water, and ΔT is the change in temperature.
How many joules of energy are necessary to heat a sample of water with a mass of 46.0 grams from 0.0 celsius to 100.0?
419.1 Joules are required to heat one gram of liquid water from 0.01 degC to 100 deg C. So the answer is 419.1*46 = 19278.6
How many joules of energy are necessary to heat a smaple of water with a mass of 46.0 grams from 0.0 c to 100.0 c?
Water has a specific heat of 4.18J/gC, so set up the specific heat equation: C (spec. heat)=q (joules)/mass x temp. change, so: 4.18 (spec. heat of water) = q/46g(100deg), so q = 4.18(4,600) = 19,228 joules (or 19.228 kJ).
A solid substance with a mass of 200 grams is at its melting point temperature in a calorimeter While the substance changes from a solid to a liquid at the same temperature the 400-gram mass of water?
The heat required for the substance to melt at its melting point is given by the formula: ( q = n \cdot \Delta H_f ), where ( q ) is the heat absorbed, ( n ) is the number of moles, and ( \Delta H_f ) is the heat of fusion. The heat released by the water during the process is: ( q = m \cdot c \cdot \Delta T ), where ( m ) is the mass, ( c ) is the specific heat capacity of water, and ( \Delta T ) is the temperature change. By equating these two equations, one can solve for the heat of fusion.
If I dissolve 50 grams of salt in 1000 mL of water and the temperature decreases from 30.5 and degC to 35.6 and degC what is the heat change enthalpy of reaction Was this an exothermic or endothermic?
How can the temperature DECREASE from 30.5ºC to 35.6ºC? That's an INCREASE in temperature. So, assuming you meant the temperature INCREASED to 35.6ºC, then it is an endothermic reaction.q = mC∆T = (1000 g)(4.184 J/g/deg)(5.1 deg) = 20,920 J = 20.9 kJ. This is the heat change for this reaction.
How much heat in kJ is required to melt 54.0 g of ice at 0 and degC into water at 0 and degC if and DeltaHfus for water 6.01 kJmol?
The first step is to convert the mass of ice to moles using the molar mass of water (18.015 g/mol). Then, use the molar enthalpy of fusion to determine the heat required to melt the ice. Finally, multiply the molar enthalpy of fusion by the number of moles of water to get the total heat required in kJ.
The specific heat of a certain type of cooking oil is 1.75 cal(g and middot and degC). How much heat energy is needed to raise the temperature of 2.40 kg of this oil from 23 and degC to 191 and degC?
The change in temperature is ΔT = 191°C - 23°C = 168°C. The heat energy Q required is given by the formula Q = mcΔT, where m is the mass, c is the specific heat, and ΔT is the temperature change. Plugging in the values: Q = 2.4 kg * 1.75 cal/(g°C) * 168°C = 7056 cal = 7.056 kcal.
What is the unit for heat?
The metric unit for heat is the calorie - the heat required to raise 1 gram of water by 1 deg C. In the SI unit system it would be the kilocalorie - the heat to raise 1 kg by 1 degC
How many joules of heat are necessary to raise the temperature of 25 g of water from 10 degrees Celsius to 60 degrees Celsius?
For an approximate calculation: specific heat capacity for water = 4.18 J/(g*degC) (how much energy is required per gram per change in degrees C) mass = 25g Change in temperature = 60-10 = 50 degC energy required = mass * change in temperature * specific heat capacity = 25g * 50 degC * 4.18 J/(g*degC)
How much heat in joules would you need to raise the temperature of 1kg of water by 5 and degC?
You would need 20,920 Joules of heat to raise the temperature of 1kg of water by 5°C. This value is calculated using the specific heat capacity of water, which is 4186 J/kg°C.
How much heat must be transferred to liquid water to change the waters temperature from 27 degrees Celsius to 32 degrees Celsius?
To change the temperature of water from 27ºC to 32ºC will depend on the mass of water that is present. Obviously, the more water, the more heat it will take. This can be calculated as follows:q = heat = mC∆T where m is the mass of water; C is sp. heat = 4.184 J/g/deg and ∆T is 5ºC (change in temp).
The specific heat of gold is 0.031 caloriesgram and degC. If 10.0 grams of gold were heated and the temperature of the sample changed by 20.0 and degC how many calories of heat energy were absorbed by?
The needed heat is:Q = 10 x 20 x 0,031 = 6,2 calories
How many calories are needed to heat 225 g of water from 50.5 degrees C to steam at 133 degrees C?
1 cal/gdegC x 225g x (100 - 50.5)degC heat to the boiling point 540 cal/g x 225g heat to vaporize the water 0.5 cal/gdegC x 225 g x (133-100)degC heat to chang temp of steam = 11137.5 cal + 121500 cal + 3712.5 cal = 14850 cal
