If you cooled 85g of water from 55c° to 2c°, what is the amount of heat lost

Q = - 18830.9 J

Explanation:

Given data:

Mass of water = 85 g

Initial temperature = 55°C

Final temperature = 2°C  

Amount of heat lost = ?

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 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 = 2°C - 55°C

ΔT = -53°C

Q = 85 g × 4.18 j/g.°C × -53°C

Q = - 18830.9 J

Negative sign shows that heat is released.

a 20-cm-long, cylindrical graphite (pure carbon) rod is inserted into an oxidizing atmosphere at 1145 k and 1.013 x 105 pa pressure. the oxidizing process is limited by the diffu- sion of oxygen counterflow to the carbon monoxide that is formed on the cylindrical surface.