Respuesta :
The question is incomplete. Here is the complete question.
Calcium oxide, [tex]CaO_{(s)}[/tex], has been proposed as a substance that can be used to heat water quickly for portable heating packs or for cooking. When placed in water, [tex]CaO_{(s)}[/tex] reacts as shown by hte equation below:
[tex]CaO_{(s)}+H_{2}O_{(l)}[/tex] ⇄ [tex]Ca(OH)_{2}_{(s)}[/tex]
a) A student wants to design a heating pad that could heat a 150.0 g sample of water from 25.0°C to 60.0°C.
(i) Calculate the amount of heat, in joules, that the water must absorb for its temperature to change by this amount. (Assume that specific heat capacity of water is 4.18J/(g.°C).)
(ii) Calculate the minimum mass of [tex]CaO_{(s)}[/tex] that the student would need to use in order to cause this temperature change.
Answer: (i) Q = 21,945J
(ii) m = 467.87g
Explanation: Energy necessary for the water to absorb to heat up is calculated by:
[tex]Q=mc\Delta T[/tex]
Q is heat
c is specific heat capacity, in this case in J/g.°C
[tex]\Delta T[/tex] is change in temperature
(i) Calculating amount of heat:
[tex]Q=150*4.18*(60-25)[/tex]
Q = 21945J
The water needs 21,945 joules to increase by that amount of temperature.
(ii) According to the chemical equation:
[tex]CaO_{(s)}+H_{2}O_{(l)}[/tex] ⇄ [tex]Ca(OH)_{2}_{(s)}[/tex]
it is needed 1 mol of each reagent to produce 1 mol of product.
Molar mass of water is 18g/mol. For 150g, there are:
n = [tex]\frac{150}{18}[/tex]
n = 8.34 mols
Molar mass of CaO is 56.1g/mol. For 8.34 mols:
m = 8.34*56.1
m = 467.87g
For calcium oxide to increase temperature in 35°C, the minimum necessary is 467.87 grams.
