Question

$500g$ of water and $100g$ of ice at $0^{o}C$ are in a calorimeter whose water equivalent is $40g$. $10g$ of steam at ${100}^{o}C$ added to it. Then water in the calorimeter is (Latent heat of ice $=80cal/g$, Latent heat of steam$=540cal/g$)

• A. $580g$
• B. $590g$
• C. $600g$
• D. $610g$

Answer 1

Answer: (B)

$590g$

$\text{Given}:$ mass of water at $0^o$ = 500g, mass of ice at $0^o$ = 100g, mass of steam at ${100}^o$ = 10g

$\text{Solution}:$

Here, the energy released by the condensation of vapour is not sufficient to melt the whole ice. So, the amount of energy released by the water vapour, when it condensed and comes to,

$E=m_{vapour}{L}_{vapour}+m_{vapour}{S}_{vapour}\times 100$

$⟹E=10\times 540+10\times 1\times 100$

$⟹E=6400cal$

So, let m mass of the ice is m'. Therefore,

m'L ice=6400

$⟹m^{\prime }=640080$

$m^{\prime }=80gm$

Therefore, the total amount of water in the capillary tube is,

$m_0=500+80+10$

$m_0=590g$

$\text{Hence B is the correct option}$