Respuesta :
Explanation:
Heat capacity is defined as the amount of heat required to raise the temperature of a substance by 1 degree celsius.
(a) It is known that specific heat of water is 4.184 [tex]J/g^{o}C[/tex].
Hence, heat capacity in joules and in calories per degree for 28.4 g of water will be as follows.
[tex]28.4 g \times 4.184 J/g^{o}C[/tex]
= [tex]119 J/^{o}C[/tex]
In calories,
[tex]28.4 g \times 1.00 cal/g^{o}C[/tex]
= 28.4 [tex]cal/^{o}C[/tex]
(b) In 1 oz there are 28.3 grams. And, heat capacity of lead is 0.129 [tex]J/g ^{o}C[/tex].
Hence, heat capacity in joules and in calories per degree for 1 oz or 28.3 g of lead will be as follows.
[tex]28.3 g \times 0.129 J/g^{o}C[/tex]
= [tex]3.65 J/^{o}C[/tex]
In calories,
[tex]28.3 g \times 0.129 J/g ^{o}C \times \frac{1 cal}{4.184 J}[/tex]
= 0.882 [tex]cal/^{o}C[/tex]
The heat capacity, in joules and in calories per degree are 118.8256 Joules and 28.38Calories per degree
The heat capacity, in joules and in calories per degree are 3.6507 Joules and 0.872Calories per degree
The formula for calculating heat capacity is expressed as:
H = mc
m is the mass of the substance
c is the specific heat capacity
a) For 28.4 g of water
H = mc
c = 4.184 J/g°C
Substitute the given parameters into the formula;
H = 28.4 * 4.184
H = 118.8256 Joules
Express in calories per degree
H = 118.8256/4.186
H = 28.38Calories per degree
The heat capacity, in joules and in calories per degree are 118.8256 Joules and 28.38Calories per degree
b) For 1.00 oz of lead (28.3)
H = mc
c for lead = 0.129 J/g°C
Substitute the given parameters into the formula;
H = 28.3 * 0.129
H = 3.6507 Joules
Express in calories per degree
H = 3.6507/4.186
H = 0.872Calories per degree
The heat capacity, in joules and in calories per degree are 3.6507 Joules and 0.872Calories per degree
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