As a 15.1-gram sample of a metal absorbs 48.75 J of heat, its temperature increases 25.0K. [tex]0.129 \frac{\mathrm{J}}{\mathrm{g} \mathrm{K}}[/tex] is the specific heat capacity of the metal
Answer: Option A
Explanation:
Specific heat term explains the amount of heat needs to be added with unit mass in order to increase the temperature by a degree Celsius. Its formula is given by,
[tex]Q=m \times c \times \Delta T[/tex]
Where,
[tex]\Delta T=\text { final } T-\text {initial} T=25.0 \mathrm{K}[/tex]
[tex]Q \text { is the heat energy in Joules }=48.75 \text { Joules }[/tex]
[tex]c \text { is the specific heat capacity }[/tex]
[tex]m \text { is the mass of the metal }[/tex]
Plugging in the values
[tex]c=\frac{Q}{m \times \Delta T}=\frac{48.75 \mathrm{J}}{15.1 \mathrm{g} \times 25.0 \mathrm{K}}=\frac{48.75 \mathrm{J}}{377.5 \mathrm{g} \mathrm{K}}=0.129 \frac{\mathrm{J}}{\mathrm{g} \mathrm{K}}(\text { Answer })[/tex]
