krit.club logo

The Particulate Nature of Matter - Specific Heat Capacity

Grade 11IBPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

•

Specific heat capacity (cc) is defined as the amount of thermal energy required to raise the temperature of a unit mass of a substance by one Kelvin (or one degree Celsius). It is measured in J kg−1 K−1\text{J kg}^{-1}\text{ K}^{-1}.

•

The change in thermal energy (QQ) is directly proportional to the mass (mm) of the substance and the change in temperature (ΔT\Delta T).

•

In the particulate nature of matter, adding thermal energy increases the average random kinetic energy of the molecules, which is macroscopically observed as an increase in temperature.

•

During a phase change, the temperature remains constant because the energy supplied is used to overcome intermolecular forces (increasing potential energy) rather than increasing the kinetic energy of particles.

•

The Principle of Conservation of Energy in calorimetry assumes that in an isolated system, the thermal energy lost by a hot object is equal to the thermal energy gained by a cooler object: Qlost=QgainQ_{\text{lost}} = Q_{\text{gain}}.

📐Formulae

Q=mcΔTQ = mc\Delta T

c=QmΔTc = \frac{Q}{m\Delta T}

ΔT=Tfinal−Tinitial\Delta T = T_{\text{final}} - T_{\text{initial}}

P=Qt=mcΔTtP = \frac{Q}{t} = \frac{mc\Delta T}{t}

💡Examples

Problem 1:

An electric heater with a power rating of 1.5 kW1.5\text{ kW} is used to heat 2.0 kg2.0\text{ kg} of water from 20∘C20^\circ\text{C} to 80∘C80^\circ\text{C}. Calculate the time required, assuming no energy is lost to the surroundings. (Specific heat capacity of water c=4180 J kg−1 K−1c = 4180\text{ J kg}^{-1}\text{ K}^{-1})

Solution:

ΔT=80−20=60 K\Delta T = 80 - 20 = 60\text{ K} Q=mcΔT=2.0×4180×60=501,600 JQ = mc\Delta T = 2.0 \times 4180 \times 60 = 501,600\text{ J} P=Qt⇒t=QPP = \frac{Q}{t} \Rightarrow t = \frac{Q}{P} t=501,6001500=334.4 st = \frac{501,600}{1500} = 334.4\text{ s}

Explanation:

First, calculate the total energy required using Q=mcΔTQ = mc\Delta T. Then, relate the power (P=1500 WP = 1500\text{ W}) to the energy using P=Q/tP = Q/t to solve for time.

Problem 2:

A 0.5 kg0.5\text{ kg} block of metal at 100∘C100^\circ\text{C} is placed into 1.0 kg1.0\text{ kg} of water at 20∘C20^\circ\text{C}. The final steady temperature of the mixture is 25∘C25^\circ\text{C}. Calculate the specific heat capacity of the metal.

Solution:

Qwater=mwcwΔTw=1.0×4180×(25−20)=20,900 JQ_{\text{water}} = m_w c_w \Delta T_w = 1.0 \times 4180 \times (25 - 20) = 20,900\text{ J} Qmetal=mmcmΔTm=0.5×cm×(100−25)=37.5×cmQ_{\text{metal}} = m_m c_m \Delta T_m = 0.5 \times c_m \times (100 - 25) = 37.5 \times c_m Qlost=Qgain⇒37.5×cm=20,900Q_{\text{lost}} = Q_{\text{gain}} \Rightarrow 37.5 \times c_m = 20,900 cm=20,90037.5≈557.3 J kg−1 K−1c_m = \frac{20,900}{37.5} \approx 557.3\text{ J kg}^{-1}\text{ K}^{-1}

Explanation:

Assuming a closed system, the heat lost by the metal equals the heat gained by the water. We calculate the heat gained by the water and set it equal to the expression for the heat lost by the metal to solve for cmc_m.

Specific Heat Capacity - Revision Notes & Key Formulas | IB Grade 11 Physics