JEE MainPhysicsModerateHeat

When 300 J of heat is added to 25 gm of sampleof a material its
temperature rises from 25ºC to45ºC. The thermal capacity of the sample
andspecific heat of the material are respectively given by

Show answer and solution

\Delta\phi = ms\Delta T

\therefore

Thermal
capacity

= ms = \frac{\Delta Q}{\Delta T} = \frac{300J}{45 - 25} = 15J/ºC

and specific
heat,

s = \frac{\Delta Q}{m\Delta T} = \frac{300}{25 \times 10^{- 3} \times (45 - 25)} = \frac{600J}{kg} - C

Source: Grass-G / P-08 Physics paper 23 October.docx

Need a video for this idea?

Scan any doubt from your notes and get a whiteboard explanation.