Physics · Thermal Properties Of Matter · NEET
Heating a whole solid sphere raises the temperature of ALL the material inside it, not just its surface. The amount of material is the mass, and mass = density times volume. The volume of a sphere is (4/3)*pi*r^3, so mass goes as r^3. Because Q = m*c*delta T and only m changes, Q also goes as r^3. Surface area (which goes as r^2) matters for radiation or heat loss from the surface, not for how much heat is stored inside.
Both spheres are the SAME material, so they have the SAME specific heat c and the SAME density rho. When you take the ratio Q1/Q2, the c cancels and the rho cancels. What is left is only the volume ratio, which is (r1/r2)^3. That is why the answer is a clean number like 27/8 with no material data needed.
Yes, that condition is essential. The rule Q proportional to r^3 only holds when both spheres are heated through the same temperature change (same delta T). If the temperature rises are different, you must keep them: Q1/Q2 = (m1*delta T1)/(m2*delta T2) = (r1/r2)^3 * (delta T1/delta T2).
Then you cannot cancel c and rho. Use the full formula Q = rho * (4/3)*pi*r^3 * c * delta T for each sphere and take the ratio. The heat ratio becomes (rho1*c1*r1^3)/(rho2*c2*r2^3) for equal delta T. The clean r^3 shortcut only works when material and delta T match.
It is about total heat capacity (the heat to raise the whole object by 1 K), which is C = m*c. Specific heat c is per kilogram and is the same for both spheres. But the heat capacity C is bigger for the bigger sphere because it has more mass. So the bigger sphere has larger heat capacity even though the specific heat is identical.
The quantities of heat required to raise the temperature of two solid copper spheres of radii r1 and r2 (r1 = 1.5 r2) through 1 K are in the ratio Q1 : Q2 equal to:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Q = m*c*delta T, where m = rho*(4/3)*pi*r^3. Combining, Q = rho*(4/3)*pi*r^3*c*delta T. For the same material and same delta T, this simplifies to Q proportional to r^3.
Heat goes as r^3, so doubling the radius (r2 = 2*r1) needs 2^3 = 8 times more heat, assuming the same material and same temperature rise.
A bigger object of the same material has more mass, so it stores more heat for each degree (its heat capacity C = m*c is larger). With the same heater power, more required heat means more time.
No. Specific heat c is a property of the material only (per kilogram) and does not depend on size or shape. Only the total heat capacity C = m*c grows with size.
When you compare surface effects like radiated power or surface heat loss, which depend on surface area (r^2). For heat stored inside a solid body, always use volume (r^3).
