Physics · Thermal Properties Of Matter · NEET
No. The simple average is correct ONLY when both bodies have the same heat capacity (same m times c). In general the final temperature is a weighted average: Tf = (m1 c1 T1 + m2 c2 T2) / (m1 c1 + m2 c2). If one body has a larger heat capacity, the final temperature is pulled closer to that body's starting temperature. This is the single most tested idea in NEET on this topic.
Because the system is isolated (no heat escapes to the surroundings). Energy is conserved, so every joule of heat that leaves the hot body must enter the cold body. We write it as m1 c1 (T1 - Tf) = m2 c2 (Tf - T2). Both sides are positive: the hot body cools from T1 down to Tf, the cold body warms from T2 up to Tf.
Specific heat c is per unit mass (unit J per kg per K). Heat capacity is the whole body's value, C = m c (unit J per K). If a question gives you heat capacities C1 and C2 directly (not masses and specific heats), just use Tf = (C1 T1 + C2 T2) / (C1 + C2). The mass has already been folded into C.
No. Because only temperature DIFFERENCES appear (T1 - Tf and Tf - T2), a difference of 1 degree Celsius equals a difference of 1 Kelvin. So you can safely keep everything in Celsius and the answer comes out in Celsius. Convert to Kelvin only if a later part of the problem needs absolute temperature (like radiation).
The calorimeter also gains or loses heat, so add its term. Heat gained by (cold water + calorimeter) = heat lost by hot body: (m_water c_water + m_cal c_cal)(Tf - T2) = m_hot c_hot (T1 - Tf). Forgetting the calorimeter term is a common mistake that shifts the answer.
Never, as long as there is no phase change (no melting or boiling) and no heat added from outside. The final temperature is a weighted average, so it must lie strictly between T2 and T1. If your calculated Tf comes out above the hot body or below the cold body, you made an algebra or sign error.
Two bodies have different thermal (heat) capacities. One of them is at 100 C and the other at 0 C. If the two are brought into contact in an isolated system (no heat loss to surroundings), the final common temperature will be:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Tf = (m1 c1 T1 + m2 c2 T2) / (m1 c1 + m2 c2). If heat capacities C = m c are given directly, use Tf = (C1 T1 + C2 T2)/(C1 + C2).
It depends on both the temperatures AND the heat capacities. The final value always lies between the two starting temperatures and leans toward the body with the larger m times c.
In an isolated system, heat lost by the hot body equals heat gained by the cold body. No heat is created or destroyed; it only moves from hot to cold until temperatures are equal.
Once both bodies reach the same temperature, there is no temperature difference to drive heat flow, so no more net heat moves. The system is in thermal equilibrium.
No. If a phase change happens, you must add latent heat terms (Q = m L) because temperature stays constant during melting or boiling. The simple weighted-average formula assumes no phase change.
