14.1 – Determining the Specific Heat Capacity of a Liquid (Leaving Cert Physics): Flashcards

Electrical energy supplied ($Q$)

Low specific heat capacity

Ensures uniform temperature throughout water

Heat loss to surroundings

Makes specific heat capacity too high

Minimises heat loss to surroundings

0.1°C

Calorimeter also gains heat during experiment

15°C

$Q = m_w c_w (θ_2 - θ_1) + m_c c_w (θ_2 - θ_1)$

System reaches stable temperature before measuring

Prevents burning and ensures even heating

Use polished calorimeter

Pre-cool water to 7°C below room temperature

Conservation of energy

Find specific heat capacity of water using electrical heating

Joulemeter

Copper has low specific heat capacity

Ensures uniform temperature throughout the water

Heat loss to the surroundings

Systematic error

Makes it too high

Calorimeter also gains heat during experiment

Approximately 15°C

$Q = m_w c_w (\theta_2 - \theta_1) + m_c c_c (\theta_2 - \theta_1)$