Given:
The heat energy added to the sample is Q = 130 J
The mass of each sample is m = 1 kg
The specific heat of aluminum is
[tex]c_A=\text{ 900 J/ kg }^{\circ}C[/tex]The specific heat of brass is
[tex]c_B\text{ = 380 J/kg }^{\circ}C[/tex]The specific heat of glass is
[tex]c_G=\text{ 840 J/kg }^{\circ}C[/tex]The specific heat of steel is
[tex]c_s=450\text{ J/kg }^{\circ}C[/tex]Required: The sample that would require the greatest temperature increase.
Explanation:
The temperature increase for a sample can be calculated by the formula
[tex]\Delta T\text{ = }\frac{Q}{mc}[/tex]On substituting the values, the temperature increase of aluminum will be
[tex]\begin{gathered} \Delta T_A=\text{ }\frac{130}{1\times900} \\ =\text{ 0.144 }^{\circ}C \end{gathered}[/tex]On substituting the values, the temperature increase of brass will be
[tex]\begin{gathered} \Delta T_B=\text{ }\frac{130}{1\times380} \\ =\text{ 0.342 }^{\circ}C \end{gathered}[/tex]On substituting the values, the temperature increase of glass will be
[tex]\begin{gathered} \Delta T_G=\text{ }\frac{130}{1\times840} \\ =\text{ 0.155 }^{\circ}C \end{gathered}[/tex]On substituting the values, the temperature increase of steel will be
[tex]\begin{gathered} \Delta Ts=\text{ }\frac{130}{1\times450} \\ =\text{ 0.289 }^{\circ}C \end{gathered}[/tex]Final Answer: The greatest temperature increase is experienced by brass.