Calorimetry — Question 14

Let,

Specific heat capacity of block P = C_p

Specific heat capacity of block Q = C_Q

From relation,

$c = \dfrac{Q}{m \times △t} \\[0.5em]$

where, c = specific heat capacity

m = mass

Q = heat energy

△t = change in temperature

Now,

$\dfrac{C_p}{C_Q} = \dfrac{\dfrac{Q}{2m × △t}}{\dfrac{Q}{m × △t}} \\[0.5em] \dfrac{C_p}{C_Q} = \dfrac{m × △t}{2m × △t} \\[0.5em] \Rightarrow \dfrac{C_p}{C_Q} = \dfrac{1}{2} \\[0.5em]$

Hence, the required ratio is 1 : 2