A sample of water drops from an initial temperature of 67.5 C to a final temperature of 25.2 C. If - 4425 J of heat were released from this sample of water, what is the mass of the sample?

mass = 25.03 g

Explanation:

Given dа ta:

Initial temperature = 67.5°C

Final temperature = 25.2°C

Heat released = - 4425 j

Mass of water = ?

Solution:

Specific heat capacity:

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

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 = 25.2°C - 67.5°C

ΔT = - 42.3°C

Specific heat of water = 4.18 j/g.°C

Q = m. c. ΔT

-4425 j = m * 4.18 j/g.°C * - 42.3°C

-4425 j = m * - 176.81 j/g

m = - 4425 j / - 176.81 j/g

m = 25.03 g