Answer: 2175.68 J heat is required to increase the temperature of 20 grams of water by 26 degrees celsius.
Explanation:
Given: Mass = 20 g
Change in temperature = [tex]26^{o}C[/tex]
The standard value of specific heat of water is [tex]4.184 J/ g^{o}C[/tex]
Formula used to calculate the heat energy is as follows.
[tex]q = m \times C \times \Delta T[/tex]
where,
q = heat energy
m = mass of substance
C = specific heat of substance
[tex]\Delta T[/tex] = change in temperature
Substitute the values into above formula as follows.
[tex]q = m \times C \times \Delta T\\= 20 g \times 4.184 J/g ^{o}C \times 26^{o}C\\= 2175.68 J[/tex]
Thus, we can conclude that 2175.68 J heat is required to increase the temperature of 20 grams of water by 26 degrees celsius.