Question

13. Base your answer to the following question on the information below and you knowledge of chemistry,
A 100.-gram sample of liquid water is heated from 20.0°C to 50.0°C. Enough KCION(S) is dissolved
in the sample of water at 50.0°C to form a saturated solution.
Using the information on Table B, determine the amount of heat absorbed by the water
when the water is heated from 20.0°C to 50.0°C.​

Answer 1

Answer:

Heat absorbed by water is 12540 J

Explanation:

Given data:

Mass of water = 100 g

Temperature changes = 20.0°C to 50.0°C

Heat absorbed by water = ?

Solution:

Specific heat of water is 4.18 j/g.°C

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

ΔT = 50.0°C - 20.0°C

ΔT = 30.0°C

Q = m.c. ΔT

Q = 100 g. 4.18 j/g.°C. 30.0°C

Q = 12540 J

Heat absorbed by water is 12540 J.