What is the amount of heat energy absorbed when 36 grams of ice at -20oC is melted to water at 0oC?

Q = 1461.6 J

Explanation:

Given data:

Mass of ice = 36 g

Initial temperature = -20°C

Final temperature = 0°C

Amount of heat absorbed = ?

Solution:

specific heat capacity of ice is 2.03 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 = T2 - T1

ΔT =  0°C - (-20°C)

ΔT = 20°C

Q = 36 g ×2.03 j/g.°C×20°C

Q = 1461.6 J

so what is the problem with your lab report and whats the subject of the lab report? i will you

the correct answer is:

a. formation of precipitation