A mysterious object with a surface area of 0.015 m2, volume of 0.000125 m3, density of 100 kg/m3, specific heat of 100 J/(kgK), thermal conductivity of 2 W/(mK), with an unknown initial temperature was placed in a fluid with a density of 50 kg/m3, specific heat of 70 J/(kgK), thermal conductivity of 0.1 W/(mK), at a temperature of 400K. The heat transfer coefficient is given to be 10 W/(m2K). After 10 seconds, the temperature of the object is measured to be 380K. Determine the object's initial temperature.

The object's initial temperature is 333.6 K

Explanation:

We first assume that the liquid can only transfer heat to the object through convective heat transfer method.

Let T₀ = the initial temperature of the object

T = temperature of the object at anytime.

The rate of heat transfer from the liquid to the object is given as

Q = -hA (T∞ - T)

T∞ = temperature of the fluid = 400 K

A = Surface area of the object in contact with the liquid = 0.015 m²

h = Convective heat transfer coefficient is given to be = 10 W/(m²K)

The rate of heat gained by the object is given by

mC (d/dt)(T∞ - T)

m = mass of the object = ρV

ρ = density of the object = 100 kg/m³

V = volume of the object = 0.000125 m³

m = ρV = 100 × 0.000125 = 0.0125 kg

C = specific heat capacity of the object = 100 J/(kgK)

The rate of heat loss by the liquid = rate of heat gain by the object

-hA (T∞ - T) = mC (d/dt)(T∞ - T)

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

- mC (dT/dt) = -hA (T∞ - T)

(dT/dt) = (hA/mC) (T∞ - T)

Let s = (hA/mC)

(dT/dt) = -s (T - T∞)

dT/(T - T∞) = -sdt

Integrating the left hand side from T₀ (the initial temperature of the object) to T and the right hand side from 0 to t

In [(T - T∞)/(T₀ - T∞)] = -st

(T - T∞)/(T₀ - T∞) = e⁻ˢᵗ

(T - T∞) = (T₀ - T∞)e⁻ˢᵗ

s = (hA/mC) = (10 × 0.015)/(0.0125×100) = 0.12

T = 380 K at t = 10 s

T₀ = ?

T∞ = 400 K

st = 0.12 × 10 = 1.2

(380 - 400) = (T₀ - 400) e⁻¹•²

(-20/0.3012) = (T₀ - 400)

(T₀ - 400) = - 66.4

T₀ = 400 - 66.4 = 333.6 K

Hope this Helps!!!

it is the second option.