Question

2. Frozen yoghurt with a temperature of 0°C is removed from a freezer and placed in a room which is at a temperature of 22°C. The temperature of the yoghurt is found to be 6°C after 5 minutes. (a) Write down an initial value problem to model the described phenomena and solve it. [6 marks] (b) Determine the temperature of the yoghurt after 10 minutes. (4 marks)

Answer 1

Answer:

A. ΔT/Δt=(3/55min)*(Tr-To); B. 12°C

Step-by-step explanation:

In order to find the answer let's detect the information.

Object temperature = To = 0°C

Room temperature = Tr = 22°C

Rate of warming = R

Based on the data we can establish that:

The change (Δ) in temperature of the object through time is proportional to the difference between object and room temperatures, so:

ΔT/Δt=R*(To-Tr)

Let's find the rate (R) first:

Because the object change its temperature from 0°C to 6°C in 5 minutes we have:

ΔT/Δt=R*(Tr-To)

(6-0°C)/5min=R*(22°C-0°C)

6/(5*22min)=R

3/55min=R

So the general equation is:

ΔT/Δt=(3/55min)*(Tr-To)

Now, let's find the temperature after 10 minutes.

ΔT/Δt=R*(Tr-To)

ΔT/10min=(3/55min)*(22°C-0°C)

ΔT=(3/55min)*(22°C)*10min

ΔT=12°C

In conclusion, after 10 minutes the yoghurt has a temperature of 12°C.