First slide

Questions on calorimetry without phase change

Question

A copper block of mass 500 gm and specific heat 0.1 cal/gm°C is heated from 30°C to 290°C, the thermal capacity of the block is

Easy

A

50cal/ºC
B

50gm
C

5cal/ºC
D

5gm

Solution

$\large \Delta Q=ms\Delta T$
$\large H=\frac {\Delta Q}{\Delta T}=ms=500(0.1)=50\;cal/^0C$

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