Question

Water is heated to 80oC for 10 min. How much would be the temperature if k = 0.056 per min and the surrounding temperature is 25oC?

Answer 1

Answer:

Explanation:

GIVEN:

Ts = 25oC,

To = 80oC,

t = 10 min,

k = 0.056

REQUIRED:

T=?

SOLUTION:

Now, substituting the above data in Newton’s law of cooling formula,

T(t) = Ts + (To – Ts) × e-kt

= 25 + (80 – 25) × e-0.56 = 25 + [55 × 0.57] = 45.6 oC

Temperature cools down from 80oC to 45.6oC after 10 min.

Answer 2

Given:

Ts = 25°C

To = 80°C

t = 10 min

k = 0.056

To find:

T=?

Solution:

In this problem, we can apply Newton's law of cooling.

According to Newton's law of cooling, the rate of loss of heat from a body is directly proportional to the difference in the temperature of the body and its surroundings.

Put all values in the above equation,

T(t) = Ts + (To - Ts)× e - kt

T(t) = 25 +(80 - 25) × e- 0.56

T(t) = 25 + [55 × 0.56]

T(t) = 45.6°c

The temperature would be 45.6°c between k= 0.056 and the surrounding temperature 25°C.