Calculate the amount of heat liberated (in kJ) from 345 g of mercury when it cools from 72.7°C to 12.0°C.

2.93181 Kj

Explanation:

Given data:

Mass of mercury = 345 g

Initial temperature = 72.7°C

Final temperature = 12.0°C

Heat released = ?

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 mercury = 0.14 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 = Final temperature - initial temperature

ΔT = 12.0°C - 72.7°C

ΔT = -60.7°C

Q = m.c. ΔT

Q = 345 g × 0.14 j/g.°C × -60.7°C

Q = -2931.81 j

Joule to Kj conversion:

2931.81 j × 1 kj /1000 j = 2.93181 Kj

its first and third is your answer i did the qustion the fourth isnt right