Find q when 22.0g of water is heated from 17.0 degrees celsius to 100 degrees celsius

When 22.0 g of water is heated from  17 °C  to 100 °C what amount of heat absorbed?

When 22.0 g of water is heated from  17 °C  to 100 °C  water absorbed   7643.64 j heat.

Explanation:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

Given data:

Mass of water = 22.0 g

Initial temperature = 17°C

Final temperature = 100°C

Solution:

ΔT  = T₂ -T₁

ΔT  = 100°C - 17 °C= 83°C

Q = m.c. ΔT

Q = 22 g× 4.186 j/g °C × 83 °C

Q = 7643.64 j

It means water require more heat which is  7643.64 j to increase its temperature from 17  °C to 100  °C .

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the dissolution of polymer can be thermodynamically described in 2 steps.