In an experiment to determine the specific latent heat of vaporization of water at 100^\ci… — JAMB UTME

In an experiment to determine the specific latent heat of vaporization of water at $100^\circ\text{C}$ using an electric immersion heater in an uninsulated vessel, if heat loss from the vessel to the cooler surrounding environment is neglected in the calculations, the experimentally determined value of the specific latent heat of vaporization will be higher than the true value.

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True

The statement is correct because the measured energy supplied by the heater includes both the energy required for the phase change and the thermal energy dissipated to the surroundings. Dividing this larger total energy by the mass of vaporized liquid results in an overestimation of the specific latent heat of vaporization.

Adım Adım Çözüm

Formulate the thermal energy balance equation including heat loss.

Q_{\text{electrical}} = Q_{\text{vaporization}} + Q_{\text{loss}}

, where

Q_{\text{electrical}} = P \cdot t

and

Q_{\text{vaporization}} = m \cdot L_{\text{true}}

Energy conservation dictates that the electrical energy supplied by the heater equals the thermal energy used for vaporization plus the heat lost to the cooler ambient surroundings.

Express the calculated specific latent heat

L_{\text{calc}}

under the assumption of zero heat loss.

L_{\text{calc}} = \frac{Q_{\text{electrical}}}{m} = \frac{m \cdot L_{\text{true}} + Q_{\text{loss}}}{m}

The experimenter measures the total electrical power and time, assuming all of this energy converted liquid into vapor.

Compare the calculated specific latent heat

L_{\text{calc}}

with the true value

L_{\text{true}}

L_{\text{calc}} = L_{\text{true}} + \frac{Q_{\text{loss}}}{m} > L_{\text{true}}

Because

Q_{\text{loss}} > 0

and mass

m > 0

, the ratio

\frac{Q_{\text{loss}}}{m}

is positive, meaning

L_{\text{calc}}

is higher than

L_{\text{true}}

Anahtar Kavram

Experimental Heat Loss Error Propagation in Latent Heat Calculations

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