Solved Examples on Heat Capacity

Example 1: A 88.3 g sample of metal at 95.24  C is added to 35.10 g of water that is initially at 17.27 °C. The final temperature of both the water and the metal is 29.20 °C. The specific heat of water is 4.184 J/(g°C). Calculate the specific heat of the metal.

Solution:

Given,

Mass of metal is 88.3 g

The initial temperature of the metal is 95.24 °C.

Mass of water: 35.10 g.

The initial temperature of the water is 17.27 °C.

The final temperature of the water and the metal is 29.20 °C.

The specific heat of water is  4.184 J/(g°C).

Therefore, the expression where the energy from the hotter metal transfers to the cooler water is

−m_oC_oΔT_o = m_wC_wΔT_w

where
m_o = mass of a metal object
ΔT_o = temperature change of metal object
C_o = specific heat capacity of metal object
m_w = mass of water
ΔT_w = temperature change of water
C_w = specific heat capacity of water

Rearrange the above expression,

C_o = (m_wC_wΔT_w)/(m_oΔT_o)

Substitute the values in the above expression,

C_o = [35.10 4.184(29.20−17.27)]/[88.3(29.20-95.24)]

    = 0.301 J/g°C

Example 2: A 30.5 g sample of an alloy at 93.0 °C is placed into 50.0 g of water at 22.0 °C in an insulated coffee cup with a heat capacity of 9.2 J/K. If the final temperature of the system is 31.1 °C, what is the specific heat capacity of the alloy?

Solution:

Heat absorbed = heat lost

then the specific heat capacity, of that substance is given by

s = (1/m)(ΔQ/ ΔT)

Rearrange the above expression,

ΔQ=smΔT

ΔQ_alloy = ΔQ_water+ΔQ_cup

Temperature of the water is equal to the temperature of the cup = 22.0 °C.

Temperature of the alloy is  93.0 °C.

Final Temperature is 31.1 °C.

30.5×(93.0 – 31.1)s = 9.2×(31.1-22.0) + 50.0×4.2×(31.1-22.0)

1887.95×s = 1994.72

s = 1.057 J/gK

Example 3: The specific heat of water is 4.18 J/(g°C). Calculate the molar heat capacity of water. Express your answer to three significant figures and include the appropriate units.

Solution:

Specific Heat of water is 4.18 J/(g°C).

Expression to convert gram into mole is

4.18 J/gC x (18.0 g / mole) = 75.24 J/mole C

Hence, the molar heat capacity of water is 75.24 J/mole C

Heat capacity or thermal capacity is an extensive property of matter, that defines its physical property. Heat Capacity is the amount of heat that must be applied to an object in order to cause a unit change in temperature. Heat capacity is measured in Joules per Kelvin (J/K), which is its SI unit. When heat capacity is divided by the mass of the substance, gives the corresponding intensive property called Specific Heat Capacity. Moreover, heat capacity divided by the amount of substance in volumes gives Molar Heat Capacity. Hence, in this article, we’ll understand the important concepts of Heat Capacity like its definition, explanation, formula, unit, specific heat capacity, molar heat capacity