The water equivalent of a calorimeter is 10 g and it contains 50 g of water at 15°C. Some amount of ice, initially at -10°C is dropped in it and half of the ice melts till equilibrium is reached. What was the initial amount of ice that was dropped (when specific heat of ice = 0.5 cal gm-1°C-1, specific heat of water = 1.0 cal gm-1°C-1 and latent heat of melting of ice = 80 cal gm-1) ?

Question

The water equivalent of a calorimeter is 10 g and it contains 50 g of water at 15°C. Some amount of ice, initially at -10°C is dropped in it and half of the ice melts till equilibrium is reached. What was the initial amount of ice that was dropped (when specific heat of ice = 0.5 cal gm-1°C-1, specific heat of water = 1.0 cal gm-1°C-1 and latent heat of melting of ice = 80 cal gm-1) ? (A) 10 g (B) 18 g (C) 20 g (D) 30 g

Tardigrade Admin · Accepted Answer

Let the mass of ice =m Applying calorimetry principle, heat given = heat taken (m1+m2) s1(t1-t)=(m L/2)+m s2(t-t2) Putting values, we get (10+50) × 1 ×(15-0)=(m/2) × 80+m × 0.5[0-(-10)] ⇒ 60 × 15=40 m+(m/2) × 10=45 m ∴ m=(60 × 15/45)=20 g

10 g

18 g

20 g

30 g

Let the mass of ice =m Applying calorimetry principle, heat given = heat taken (m1​+m2​)s1​(t1​−t)=2mL​+ms2​(t−t2​) Putting values, we get (10+50)×1×(15−0)=2m​×80+m×0.5[0−(−10)] ⇒60×15=40m+2m​×10=45m ∴m=4560×15​=20g