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.
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.
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 ).
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.
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.
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.
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
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).
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.
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.
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 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.
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
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)
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.
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 needed heat is:Q = 10 x 20 x 0,031 = 6,2 calories
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