Respuesta :
Answer:
ΔT ∝ [tex]\frac{1}{m}[/tex]
Explanation:
We know that, the heat input to a body is given as:
[tex]Q_{in}=mc\Delta T[/tex]
Where,
[tex]Q_{in} \to \textrm{Heat in}\\m\to \textrm{mass of the body}\\c\to \textrm{Specific heat capacity of the body}\\\Delta T\to\textrm{Temperature rise of the body}[/tex]
Specific heat capacity of a body depends on its material and thus is constant for a given body.
Rewriting the above equation in terms of [tex]\Delta T[/tex], we get:
[tex]\Delta T=\frac{Q_{in}}{mc}[/tex]
Now, as per given question, the heat supply to the given body is a constant.
Therefore, replacing the constant quantities by a constant 'K', we get:
[tex]\Delta T=\frac{Q_{in}}{c}\times \frac{1}{m}\\\\\Delta T=\frac{K}{m}[/tex]
So, rise in temperature is a function of the mass only and varies inversely with the mass.
⇒ ΔT ∝ [tex]\frac{1}{m}[/tex]
Therefore, the temperature rise of a body is INVERSELY PROPORTIONAL to its mass for the same heat energy input.
