Question

A liquid at temperature 7575F is placed in an oven at temperature 450450. The temperature of the liquid increases at a rate 77 times the difference between the temperature of the liquid and that of the oven. Write a differential equation for the temperature T(t) of the liquid.

Answer 1

The differential equation representing the temperature of the liquid as a function of time is \(\frac{dT}{dt} = 7(450 - T)\), where T is the temperature of the liquid and t is time.

Step-by-step explanation:

To construct a differential equation for the temperature T(t) of the liquid, let's consider the information provided. The temperature of the liquid increases at a rate seven times the difference between the temperature of the liquid (T) and that of the oven (450°F). This can be expressed mathematically as:

\[\frac{dT}{dt} = 7(T_{oven} - T)\]

Since the oven temperature is constant at 450°F, the equation simplifies to:

\[\frac{dT}{dt} = 7(450 - T)\]

Which is the desired differential equation, where T is the temperature of the liquid and t represents time in minutes, hours, or any other time unit consistent with the rate of change.

Answer 2

Answer:
`(dT(t))/(dt)=77* (450-T)`

Step-by-step explanation:

Given

The temperature of the liquid is
`75^(\circ)F` placed in an oven with temperature of
`450^(\circ)F`.

Initially difference in temperature of the two

`\Delta T=450-75\\\Rightarrow \Delta T=375^(\circ)F`

According to the question

`\Rightarrow (dT(t))/(dt)=77\cdot \Delta T\\\\\Rightarrow (dT(t))/(dt)=77* (450-T)\quad [\text{T=75}^(\circ)F\ \text{at t=0}]`