Consider a small but known volume of metal that has a large thermal conductivity. (a) Since the thermal conductivity is large, spatial temperature gradients that develop within the metal in response to mild heating are small. Neglecting spatial temperature gradients, derive a differential equation that could be solved for the temperature of the metal versus time T(t) if the metal is subjected to a fixed surface heat rate q supplied by an electric heater. (b) A student proposes to identify the unknown metal by comparing measured and predicted thermal responses. Once a match is made, relevant thermophysical properties might be determined, and, in turn, the metal may be identified by comparison to published property data. Will this approach work? Consider aluminum, gold, and silver as the candidate metals.

a) dT/dt = q / p*V*c_p

b) See explanation

Explanation:

Given:

- Small but knon volume of metal has large thermal conductivity.

- Electrical heater supplies energy at of q.

Find:

- The expression for temperature T with respect to time t

Solution:

- Set up an energy balance

                               E_in = E_st

                               q = m*c_p*dT/dt  

Since,volume is known we can re-write m as:

                               q = p*V*c_p*dT/dt

                              dT/dt = q / p*V*c_p

Where,

p: The density of metal

V: Volume of metal

cp: The constant pressure specific heat capacity of metal.

b) Aluminium, Gold, and silver properties from thermodynamic table are as follows:

   Aluminium: c_p = 903 J/kgK    ,     p = 2702 kg/m^3

   Gold:           c_p = 129 J/kgK     ,     p = 19,300 kg/m^3

   Silver:         c_p = 235 J/kgK     ,     p = 10,500 kg/m^3

Using the energy equation derived above dT/dt would give us similar results since the difference in p and c_p product ( p*c_p ) between the metals is small to be measured by available instruments. This approach might work for another person/candidate but not for the one given in part A.

a lot of it means it can be "burned" for energy and sold at one price. if it is scarce, the price will go up

f=ma newton

f=100x2.5=250n