Question

A new alloy has a density of 7380 kg/m3. Its specific heat has not yet been measured. The same new alloy of the last exercise has a melting point of 1540 K. Its thermal expansion coefï¬cient has not yet been measured. How would you make an approximate estimate of its value? What value would you then report?

Answer 1

The value of specific heat is
`271`
`(J)/(Kg.K)` and coefficient of thermal expansion is
`1.29* 10^(-5)`

Step-by-step explanation:

Given:

Density of alloy
`\rho = 7380`
`(kg)/(m^(3) )`

For finding the specific heat of alloy we use formula of specific heat in case of solid material,

`\rho C_(p) = 2 * 10^(6)`
`(J)/(m^(3). K )`

Where
`\rho =` density of alloy,

`C_(p) = (2 * 10^(6) )/(7380)`

`C_(p)= 271`
`(J)/(Kg.K)`

For finding the coefficient of thermal expansion,

`\alpha T = 0.02`

Where
`\alpha =` coefficient of thermal expansion

`\alpha = (0.02)/(1540)`

`\alpha = 1.29 * 10^(-5)`

Therefore, the value of specific heat is
`271`
`(J)/(Kg.K)` and coefficient of thermal expansion is
`1.29* 10^(-5)`