Answer:
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!!!
