Question

An object of 3 kg mass absorbs 600 cal heat and its temperature rises from 10°C to 70°C. What its specific heat ?​

Answer 1

GiveN :-

• An object of 3 kg mass absorbs 600 cal heat .
• Its temperature rises from 10°C to 70°C.

To FinD :-

• Its Specific Heat .

SolutioN :-

We are given here that a object of mass 3kg absorbs 600 calories of heat & its temperature rises from 10°C to 70°C . We need to find the specific heat . We know that (Q) = ms∆T .

Where ,

$\red{\frak{Where }}\begin{cases} \textsf{ Mass is \textbf{m} .}\\\textsf{ Specific Heat is \textbf{s} .}\\\textsf{ Heat absorbed is \textbf{Q}.}\\\textbf{ Change in temperature is \Delta \textbf{T}.}\end{cases}$

$\red{\bigstar}\underline{\textsf{ Put on the respective values :- }}$

$\sf\dashrightarrow \pink{ Q = ms\Delta T }\\\\\sf\dashrightarrow 600 cal = (3000 g)(s)(70^oC - 10^o C ) \\\\\sf\dashrightarrow S =\dfrac{ 600 }{3000\times 10 }\\\\\sf\dashrightarrow \underset{\blue{\sf Required \ Specific\ Heat }}{\underbrace{\boxed{\pink{\frak{ Specific \ Heat \ (s) \ = 0.02 \ cal \ g^{-1} \ ^{\circ} C ^{-1} }}}}}$

Answer 2

Answer:

We know that :-

=> Q = ms∆T

where , Q is heat , m is mass , s is specific heat and ∆T change in temperature .

• ∆T = 70°C - 10°C = 60°C
• Mass = 3kg=3000 g

Substitute respective values ,

=> 600 = 3000 * s * 60° C

=> s = 600 / 3000 * 60

=> s = 0.002 cal / g- °C