A metallic sphere cools from 50 ∘ C to 40 ∘ in 300 s. If atmospheric temperature around is 20 ∘ C , then the sphere's temperature after the next 5 minutes will be close to:

50 - 40 300 = - k 50 + 40 2 - 20 40 - T 300 - k T + 40 2 - 20 ⇒ 10 40 - T = 25 T * 2 ⇒ T = 200 - 5 T ⇒ T = 100 3 ≈ 33 ? C

A metallic sphere cools from $50^{\circ} C$ to $40^{\circ}$ in 300 s. If atmospheric temperature around is $20^{\circ} C$ , then the sphere's temperature after the next 5 minutes will be close to:

Option 1 - <math> <msup> <mrow> <mrow> <mn>35</mn> </mrow> </mrow> <mrow> <mrow> <mo>∘</mo> </mrow> </mrow> </msup> <mi mathvariant="normal">C</mi> </math>

Option 2 - <math> <msup> <mrow> <mrow> <mn>31</mn> </mrow> </mrow> <mrow> <mrow> <mo>∘</mo> </mrow> </mrow> </msup> <mi mathvariant="normal">C</mi> </math>

Option 3 - <math> <msup> <mrow> <mrow> <mn>33</mn> </mrow> </mrow> <mrow> <mrow> <mo>∘</mo> </mrow> </mrow> </msup> <mi mathvariant="normal">C</mi> </math>

Option 4 - <math> <msup> <mrow> <mrow> <mn>28</mn> </mrow> </mrow> <mrow> <mrow> <mo>∘</mo> </mrow> </mrow> </msup> <mi mathvariant="normal">C</mi> </math>

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Answered by

Alok Kumar Singh | Contributor-Level 10

6 months ago

Correct Option - 3

Detailed Solution:

$\frac{50 - 40}{300} = - k [\frac{50 + 40}{2} - 20]$

$\begin{array}{r} \frac{40 - T}{300} - k [\frac{T + 40}{2} - 20] \\ \Rightarrow \frac{10}{40 - T} = \frac{25}{T} * 2 \\ \Rightarrow T = 200 - 5 T \\ \Rightarrow T = \frac{100}{3} \approx 33^{?} C \end{array}$