Answer:
The samples specific heat is 14.8 J/kg.K
Explanation:
Given that,
Weight = 28.4 N
Suppose, heat energy [tex]E=1.25\times10^{4}\ J[/tex]
Temperature = 18°C
We need to calculate the samples specific heat
Using formula of specific heat
[tex]Q=mc\Delta T[/tex]
[tex]c=\dfrac{Q}{m\Delta T}[/tex]
Where, m = mass
c = specific heat
[tex]\Delta T[/tex] = temperature
Q = heat
Put the value into the formula
[tex]c=\dfrac{1.25\times10^{4}}{\dfrac{28.4}{9.8}\times(18+273)}[/tex]
[tex]c=14.8\ J/kg. K[/tex]
Hence, The samples specific heat is 14.8 J/kg.K