Question

A piece of unknown material of mass 0.15 kg is heated to 114°C without melting and then placed in a calorimeter containing 0.74 kg of water initially at 14.9°C. The system reaches an equilibrium temperature of 20.2°C. what is the specific heat of the unknown material, in units of J (kg·°C)? The specific heat of water is 4.186 x 103 J (kg·℃

Answer 1

To perform this exercise it is necessary to use the theory of the first Law of Thermodynamics, in relation to Heat Capacity. If one adds heat to an object, its temperature usually increases (exceptions include at a state boundary, for example when a liquid boils). In many cases the temperature change is linear in the amount of heat added. We define the heat capacity C of an object from the relation:

`\Delta Q = C \Delta T`

`\Delta Q = mc\Delta T`

Where

`\Delta Q =` Heat that flows into system

`\Delta T =`Temperature

c = Specific heat

m = Mass

From our values we have the exchange of heat of the material and of the water, that is,

For the water:

`\Delta Q_1 = mc_1 (T-T_2)`

Where,

`m = 0.74Kg`

`T = 20.2\°C`

`T_2 = 14.9\°C`

`c_1 = 4.186*10^3J/Kg\°C`

For the Material:

`\Delta Q_2 = mc_2 (T_1-T)`

Where,

`m = 0.15Kg`

`T = 20.2\°C`

`T_1 = 114\°C`

`c_2 = ?`

For conservation of energy in balance we have to,

`\Delta Q_1 = \Delta Q_2`

`mc_1 (T-T_2) = mc_2 (T_1-T)`

`(0.74)(4.186*10^3)(20.2-14.7)=(0.15)(c_2)(114-20.2)`

`c_2 = ((0.74)(4.186*10^3)(20.2-14.7))/((0.15)(114-20.2))`

`c_2 = 1210.8756J/Kg\°C`