Question

A cold drink initially at 39°F warms up to 43°F in 2 min while sitting in a room of temperature 72°F. How warm will the drink be if left out for 20 min?

If the drink is left out for 20 min, it will be about__________°F

Answer 1

If the cold drink is left out for 20 minutes, it will be approximately 43.4°F.

Step-by-step explanation:

To determine how warm the drink will be if left out for 20 minutes, we can use the principle of heat transfer. The rate of temperature change can be calculated using the formula Q = mc∆T, where Q is the amount of heat transferred, m is the mass, c is the specific heat capacity, and ∆T is the change in temperature.

In this case, we can assume the drink has the same specific heat capacity as water, which is 1 calorie/gram°C. The mass of the drink is not provided, so let's assume it is 250 grams (roughly the volume of the drink).

Using the formula and the given information, the amount of heat transferred in 2 minutes can be calculated: Q = (250 g) x (1 cal/gram°C) x (43°F - 39°F) = 100 calories. This means that 100 calories of heat were transferred to the drink in 2 minutes.

If the same rate of heat transfer continues, the amount of heat transferred in 20 minutes can be calculated: Q = (250 g) x (1 cal/gram°C) x (∆T). Solving for ∆T: ∆T = Q / (250 g x 1 cal/gram°C) = 100 calories / 250 g = 0.4°C.

Since the initial temperature was 43°F, the final temperature after 20 minutes will be 43°F + 0.4°C = 43.4°F. So, if the drink is left out for 20 minutes, it will be approximately 43.4°F.