An object is removed from a room where the temperature is 69 degrees and is taken outside, where the air temperature is 30 degrees. After 1 minute, the temperature of the object reads 52 degrees. What will be the temperature of the object at t

The temperature of the object at any time t, T(t) is given as

T = T∞ + (T₀ - T∞)e⁻⁰•⁵⁷²⁵ᵗ

Explanation:

Let T be the temperature of the object at any time

T∞ be the temperature outside = 30°

T₀ be the initial temperature of the object in the room = 69°

And m, c, h are all constants from the cooling law relation

From Newton's law of cooling

Rate of Heat loss by the object = Rate of Heat gain by the outside air

- mc (d/dt)(T - T∞) = h (T - T∞)

(d/dt) (T - T∞) = dT/dt (Because T∞ is a constant)

dT/dt = (-h/mc) (T - T∞)

Let (h/mc) be k

dT/(T - T∞) = -kdt

Integrating the left hand side from T₀ to T and the right hand side from 0 to t

In [(T - T∞)/(T₀ - T∞)] = -kt

(T - T∞)/(T₀ - T∞) = e⁻ᵏᵗ

(T - T∞) = (T₀ - T∞)e⁻ᵏᵗ

Inserting the known variables

(T - 30) = (69 - 30)e⁻ᵏᵗ

(T - 30) = 39 e⁻ᵏᵗ

At 1 minute, T = 52°

52 - 30 = 39 e⁻ᵏᵗ

22/39 = e⁻ᵏᵗ

- kt = In (22/39) = In (0.564)

- k(1) = - 0.5725

k = 0.5725 /min

(T - T∞) = (T₀ - T∞)e⁻⁰•⁵⁷²⁵ᵗ

T = T∞ + (T₀ - T∞)e⁻⁰•⁵⁷²⁵ᵗ

show the diagram then i will answer

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