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NEET PHYSICSThermal Properties of MatterMedium

Question

Two identical bodies are made of a material for which the heat capacity increases with temperature. One of these is at 100C100^{\circ}\text{C}, while the other one is at 0C0^{\circ}\text{C}. If the two bodies are brought into contact, then assuming no heat loss, the final common temperature is -

A

50C50^{\circ}\text{C}

B

more than 50C50^{\circ}\text{C}

C

less than 50C50^{\circ}\text{C} but greater than 0C0^{\circ}\text{C}

D

0C0^{\circ}\text{C}

Step-by-Step Solution

Let the final common temperature be TfT_f. According to the principle of calorimetry, assuming no heat loss to the surroundings, the heat lost by the hot body is equal to the heat gained by the cold body. Tf100C(T)dt=0TfC(T)dt\int_{T_f}^{100} C(T) dt = \int_{0}^{T_f} C(T) dt Since the heat capacity C(T)C(T) increases with temperature, the average heat capacity of the hot body (between TfT_f and 100C100^{\circ}\text{C}) is greater than the average heat capacity of the cold body (between 0C0^{\circ}\text{C} and TfT_f). For the two integrals to be equal, the temperature interval for the hot body must be smaller than the temperature interval for the cold body. (100Tf)<(Tf0)(100 - T_f) < (T_f - 0) 100<2Tf100 < 2T_f Tf>50CT_f > 50^{\circ}\text{C} Thus, the final common temperature will be more than 50C50^{\circ}\text{C}.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from Thermal Properties of Matter. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSThermal Properties of Matteridenticalbodiesmaterialcapacityincreases

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