At 1 atm, how much energy is required to heat 75.0 g of H2O(s) at –22.0 °C to H2O(g) at 145.0 °C? Helpful constants can be found here.
Quantity | per gram | per mole |
Enthalpy of fusion | 333.6 J/g | 6010. J/mol |
Enthalpy of vaporization | 2257 J/g | 40660 J/mol |
Specific heat of solid H2O (ice) | 2.087 J/(g·°C) * | 37.60 J/(mol·°C) * |
Specific heat of liquid H2O (water) | 4.184 J/(g·°C) * | 75.37 J/(mol·°C) * |
Specific heat of gaseous H2O (steam) | 2.000 J/(g·°C) * | 36.03 J/(mol·°C) * |
Enthalpy of the reaction is equal to the amount of energy absorbed in a reaction.
Calculate the amount of heat required by using the following formula.
Here, q is heat released by the solution that equal to . Here, n indicates that number of moles. M is mass of the sample, is the specific heat, and is temperature change.
The amount of heat required to raise one degree of temperature of one unit mass of sample is called as specific heat.
Calculate the energy in the form of heat as shown below.
The enthalpy of the change is as follows:
The enthalpy of the change is as follows:
The enthalpy of the vaporization is as follows:
The enthalpy of the change is as follows:
The enthalpy of the change is as follows:
Ans:
Energy required to heat the 75.0 g of water is .
At 1 atm, how much energy is required to heat 75.0 g of H2O(s) at –22.0...
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