5 kg of ice at 0°C is mixed with 20 kg of water at 15°C. The final temperature of mixture is (Given: specific heat of water = 1 cal g-1 °C-1, latent heat of fusion of ice = 80 cal g-1)

Question

5 kg of ice at 0°C is mixed with 20 kg of water at 15°C. The final temperature of mixture is (Given: specific heat of water = 1 cal g-1 °C-1, latent heat of fusion of ice = 80 cal g-1) (A) -4°C (B) 0°C (C) 4°C (D) 3°C

Tardigrade Admin · Accepted Answer

Let T is temperature of mixture. Heat lost by water = Heat gained by ice to melt + Heat gained by water formed from ice 20 × 1 (15 - T) = 5 × 80 + 5 × 1 ( T - 0) 300 - 20T = 400 + 5T -25 T = 100 T = -4°C

Q. $5 k g$ of ice at $0^{\circ} C$ is mixed with $20 k g$ of water at $1 5^{\circ} C$ . The final temperature of mixture is (Given : specific heat of water $= 1 c a l g^{- 1}^{\circ} C^{- 1}$ , latent heat of fusion of ice $= 80 c a l g^{- 1}$ )

$- 4^{\circ} C$

$0^{\circ} C$

$4^{\circ} C$

$3^{\circ} C$

Let $T$ is temperature of mixture.
Heat lost by water $=$ Heat gained by ice to melt $+$ Heat gained by water formed from ice
$20 \times 1 (15 - T) = 5 \times 80 + 5 \times 1 (T - 0)$
$300 - 20 T = 400 + 5 T$
$- 25 T = 100$
$T = - 4^{\circ} C$

Q. 5kg of ice at 0∘C is mixed with 20kg of water at 15∘C. The final temperature of mixture is (Given : specific heat of water =1calg−1∘C−1, latent heat of fusion of ice =80calg−1)

−4∘C

0∘C

4∘C

3∘C