On a winter day the temperature drops from -15°C to -25°C overnight. If a pan sitting outside contains 0.50 kg of ice, how much heat is removed from the ice for this temperature change?
A) 0.25 kcal
B) 2.5 kcal
C) 10 kcal
D) 1 kcal

Question

Quizplus · Accepted Answer

The heat removed from the ice can be calculated using the formula:Q = m * L * (T2 - T1)where:m is the mass of the ice (0.50 kg)L is the latent heat of fusion of ice (334 kJ/kg)T2 is the final temperature (-25°C)T1 is the initial temperature (-15°C)Plugging in the values, we get:Q = 0.50 kg * 334 kJ/kg * (-25°C - (-15°C))Q = 2.5 kcalTherefore, the correct answer is B) 2.5 kcal.

On a Winter Day the Temperature Drops from -15°C to -25°C