Answer:
If 1,079.75 Joules of heat are added to 77.75 grams of water, by 3.32 degrees Celsius the temperature of water will increase
[tex]\Delta T=3.32 C^{0}[/tex]
Explanation:
[tex]q = mC\Delta T[/tex]
Here , q = heat added / removed from the substance
m = mass of the substance taken
[tex]\Delta T[/tex] = Change in temperature
C = specific heat capacity of the substance
In liquid state the value of C for water is :
[tex]4.18 J/gC^{0}[/tex]
Given values :
q = 1079.75 J
m = 77.75 gram
Insert the value of C, m , q in the given equation
[tex]q = mC\Delta T[/tex]
on transposing ,
[tex]\Delta T=\frac{q}{mC}[/tex]
[tex]\Delta T=\frac{1079.75}{77.75\times 4.18}[/tex]
[tex]\Delta T=3.32 C^{0}[/tex]
