Question

A heater, which provides thermal energy at the rate of 48 W, is placed in some crushed ice that has been placed in a funnel. The heater is switched on for 200 s and 32 g of ice are found to have melted during this time. The excessive water is drained out through the funnel. Calculate the specific latent heat of fusion for H²O.

Answer 1

Answer on Question 57107, Physics, Other

Question:

A heater, which provides thermal energy at the rate of $48\,W$, is placed in some crushed ice that has been placed in a funnel. The heater is switched on for $200\,s$ and $32\,g$ of ice are found to have melted during this time. The excessive water is drained out through the funnel. Calculate the specific latent heat of fusion for water.

Solution:

Let's write the formula for the latent heat required to change the state from ice at $0{}^{\circ}\mathrm{C}$ to water at $0{}^{\circ}\mathrm{C}$:

$$Q = m L_f,$$

here, $m$ is the mass and $L_f$ is the specific latent heat of fusion for water.

From the other hand, we can write the formula for amount of heat that a heater provides at the rate of $48\,W$ during $200\,s$:

$$Q = P t.$$

Then, we can equate both expressions and find the specific latent heat of fusion for water:

$$m L_f = P t,$$ $$L_f = \frac{P t}{m} = \frac{48\,W \cdot 200\,s}{32 \cdot 10^{-3} k g} = 3.0 \cdot 10^5 \frac{J}{k g}.$$

Answer:

The specific latent heat of fusion for water is $L_f = 3.0 \cdot 10^5 \frac{J}{kg}$.