Question

A calculator is required to obtain the final answer on this question. A solid metal sphere at room temperature 20oC is dropped into a container of boiling water (100oC). If the temperature of the sphere increases 15o in 9 seconds, find the temperature of the ball after 18 seconds in the boiling water. (Assume the sphere obeys Newton's Law of Cooling.)

Answer 1

Answer:

T = 47.1875°C

Step-by-step explanation:

Given:

Surrounding temp, Ts = 100°C

Initial Temperature ,T0= 20°C

Increase in temperature = 15°C

Final temperature, T = 20 + 15 = 35°C

Time, t = 9 seconds

Let's take Newton's law of cooling:

$T - Ts = (T_0 - Ts)e^k^t$

We'll solve for k

$35 - 100 = (20 - 100)e^9^k$

$-65 = (-80)e^9^k$

Divide both sides by -5

$13 = (16)e^9^k$

$9k = ln (\frac{13}{16})$

$k = \frac{1}{9} ln (\frac{13}{16})$

$k = -0.02307104053$

Let's now find the temperature of the ball after 18 seconds in boiling water.

Use the Newton's equation again:

$T - Ts = (T_0 - Ts)e^k^t$

$T - 100 = (20 - 100)e^-0.0^2^3^0^7^1^0^4^0^5^3^*^1^8$

$T - 100 = (-80)e^-0.0^2^3^0^7^1^0^4^0^5^3^*^1^8$

$T - 100= -52.8125$

$T = 100 - 52.8125$

T = 47.1875°C

Temperature of the ball after 18 seconds in boiling water is 47.1875°C