How many joules are needed to change the temperate of 22g of water from 18°C to 33°C?

Q =  1379.4 J

Explanation:

Given data:

Mass of water = 22  g

Initial temperature = 18°C

Final temperature = 33°C

Heat absorbed = ?

Solution:

Specific heat capacity:

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

Specific heat capacity 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 = 33°C - 18 °C

ΔT =  15°C

Q = 522 g ×4.18 J/g.°C× 15°C

Q =  1379.4 J

89.7%

Explanation:

Percentage yield is the ratio of the actual yield of a product to the theoretical yield of a product  expressed as a percentage.

Percentage yield =  x 100

Given,

Actual yield of a product = 2.80 g

Theoretical yield of a product = 3.12 g

Percentage yield =  x 100 = 89.7 %