You visited us 1 times! Enjoying our articles? Unlock Full Access!

Evaporation

A vessel cont...

Question

A vessel contains ice and is in thermal equilibrium at (-10)°C and is supplied heat energy at the rate of 20 cal s^-1 for 450 seconds. If the mass of ice is 0.1 kg and due to the supply of heat energy, the whole ice just melts. Find the water equivalent of the vessel.
(take specific heat of ice = 0.5 cal g^-1 °C^-1 and specific heat of the vessel is 0.1 cal g^-1 °C^-1. Latent heat of fusion = 80 cal g^-1 and assume no heat is transferred to the surroundings)

Open in App

Solution

Dear student,
Find the answer to your query as under:

$T h e a m o u n t o f h e a t a b s o r b e d b y t h e i c e a t - 10^{o} C t o j u s t m e l t = Q_{1} = m_{1} C_{1} ∆ T + m_{1} L Q_{1} = 0.1 \times 0.5 \times 10 + 0.1 \times 80 Q_{1} = 40.5 C a l N o w, T o t a l a m o u n t o f s u p p l i e d = Q = R a t e \times t i m e = 20 \times 450 = 9000 C a l A m o u n t o f h e a t a b s o r b e d b y t h e v e s s e l = Q_{2} = Q - Q_{1} = 9000 - 40.5 = 8959.5 a l s o, Q_{2} = m_{2} C_{2} ∆ T w a t e r e q u i v a l e n t = M = \frac{Q_{2}}{∆ T} = \frac{8959.5}{10} = 895.95 C a l /^{o} C$

Regards
Satyendra Singh

Suggest Corrections

Similar questions

- 10^{\circ} C

150 g

20^{\circ} C

46 g

= 1 cal/g^{\circ} C

= 0.5 cal/g^{\circ} C

= 80 cal/g

2 k g

- 20^{\circ} C

H_{2} O

20^{\circ} C

1 k c a l k g^{- 1 \circ} C^{- 1}

0.5 k c a l k g^{- 1 \circ} C^{- 1}

80 k c a l k g^{- 1}

- 10^{\circ} C

50 c a l s^{- 1}

50^{\circ} C

0.2 c a l g^{- 1}^{\circ} C^{- 1}

\frac{1}{2} c a l g^{- 1}^{\circ} C^{- 1}

80 c a l g^{- 1}

1 c a l g^{- 1}^{\circ} C^{- 1}

2

- 20^{o} C

5

20^{o} C

0.5 c a l / g^{C}

1 c a l / g^{C}

80 c a l / g

100^{\circ} C

- 10^{\circ} C

= 0.5 c a l g^{- 1}^{\circ} C^{- 1}

= 540 c a l g^{- 1},

= 80 c a l g^{- 1}

Join BYJU'S Learning Program