Question

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Answer 1

Answer:

• Temperature (T) of The Body will be 28°C.

Given:

Case-1

1. Initial Temperature (T₁) = 60°C
2. Final Temperature (T₂) = 40°C
3. Surroundings Temperature (T₀) = 10°C
4. Time taken (t) = 7 min

Case-2

1. Initial Temperature (T₁) = 40°C
2. Final Temperature (T₂) = T°C
3. Surroundings Temperature (T₀) = 10°C
4. Time taken (t) = 7 min

Explanation:

$\rule{300}{1.5}$

From Newton's Law of Cooling.

$\large\bigstar \: {\boxed{\tt ln\Bigg(\dfrac{T_1 - T_0}{T_2 - T_0}\Bigg) = kt}}$

$\bold{Here}\begin{cases}T_1 \: \text{Denotes Initial Temperature} \\ T_2 \: \text{Denotes Final Temperature} \\ T_0 \: \text{Denotes SurroundingTemperature} \\ \text{t Denotes Time taken} \\ \text{k is Constant}\end{cases}$

Now,

$\large{\boxed{\tt ln\Bigg(\dfrac{T_1 - T_0}{T_2 - T_0}\Bigg) = kt}}$

Substituting the values,

$\large{\tt \longmapsto ln\Bigg(\dfrac{60 - 10}{40 - 10}\Bigg) = k \times 7}$

$\large{\tt \longmapsto ln\Bigg(\dfrac{50}{30}\Bigg) = k \times 7}$

$\large{\tt \longmapsto ln\Bigg(\cancel{\dfrac{50}{30}}\Bigg) = 7k}$

$\large{\tt \longmapsto ln\Bigg(\dfrac{5}{3}\Bigg) = 7k}$

$\large{\tt \longmapsto ln\Bigg(\dfrac{5}{3}\Bigg) = 7k \: -----(1)}$

$\rule{300}{1.5}$

$\rule{300}{1.5}$

From Newton's Law of Cooling.

$\large\bigstar \: {\boxed{\tt ln\Bigg(\dfrac{T_1 - T_0}{T_2 - T_0}\Bigg) = kt}}$

$\bold{Here}\begin{cases}T_1 \: \text{Denotes Initial Temperature} \\ T_2 \: \text{Denotes Final Temperature} \\ T_0 \: \text{Denotes SurroundingTemperature} \\ \text{t Denotes Time taken} \\ \text{k is Constant}\end{cases}$

Now,

$\large{\boxed{\tt ln\Bigg(\dfrac{T_1 - T_0}{T_2 - T_0}\Bigg) = kt}}$

Substituting the values,

$\large{\tt \longmapsto ln\Bigg(\dfrac{40 - 10}{T - 10}\Bigg) = k \times 7}$

$\large{\tt \longmapsto ln\Bigg(\dfrac{30}{T - 10}\Bigg) = 7k}$

$\large{\tt \longmapsto ln\Bigg(\dfrac{30}{T - 10}\Bigg) = 7k \: -----(2)}$

Comparing Equation (1) & Equation (2).

$\large{\tt \longmapsto ln\Bigg(\dfrac{5}{3}\Bigg) = ln\Bigg(\dfrac{30}{T - 10}\Bigg)}$

$\large{\tt \longmapsto \dfrac{5}{3} = \dfrac{30}{T - 10}}$

Now,

$\large{\tt \longmapsto 5(T - 10) = 3 \times 30}$

$\large{\tt \longmapsto 5T - 50 = 90}$

$\large{\tt \longmapsto 5T = 90 + 50}$

$\large{\tt \longmapsto 5T = 140}$

$\large{\tt \longmapsto T = \dfrac{140}{5}}$

$\large{\tt \longmapsto T = \cancel{\dfrac{140}{5}}}$

$\huge{\boxed{\boxed{\tt T = 28 \: \degree C}}}$

∴ Temperature (T) of The Body will be 28°C (Option-A).

$\rule{300}{1.5}$

Answer 2

Answer:

The answer is 28c

hope its helpful dear friend