10 g of ice at 0°C is mixed with 100 g of water at 50°C in a calorimeter. The final temperature of the mixture is [Specific heat of water = 1 cal g-1 °C-1, latent heat of fusion of ice = 80 cal g-1]

Question

10 g of ice at 0°C is mixed with 100 g of water at 50°C in a calorimeter. The final temperature of the mixture is [Specific heat of water = 1 cal g-1 °C-1, latent heat of fusion of ice = 80 cal g-1] (A) 31.2°C (B) 32.8°C (C) 36.7°C (D) 38.2°C

Tardigrade Admin · Accepted Answer

Here, Mass of water, mw = 100 g Mass of ice, mi= 10 g Specific heat of water, sw=1 cal g-1 °C-1 Latent heat of fusion of ice, Lfi = 80 cal g-1 Let T he the final temperature of the mixture. Amount of heat lost by water =mwsw(Δ T)w=100 × 1 ×(50-T) Amount of heat lost by water =miLfi+misw(Δ T)i =10 × 80 +10 ×1×(T-0) According to principle of calorimetry Heat lost = Heat gained 100 × 1 × (50 - T) = 1 0 × 8 0 + 10 × 1 × (T -0) 500 - 10T = 80 + T 11T = 420 or 7 =38.2 °C

Q. $10 g$ of ice at $0^{\circ} C$ is mixed with $100 g$ of water at $5 0^{\circ} C$ in a calorimeter. The final temperature of the mixture is [Specific heat of water $= 1 c a l g^{- 1}^{\circ} C^{- 1}$ , latent heat of fusion of ice $= 80 c a l g^{- 1}$ ]

$31. 2^{\circ} C$

$32. 8^{\circ} C$

$36. 7^{\circ} C$

$38. 2^{\circ} C$

Q. 10g of ice at 0∘C is mixed with 100g of water at 50∘C in a calorimeter. The final temperature of the mixture is [Specific heat of water =1calg−1∘C−1, latent heat of fusion of ice =80calg−1]

31.2∘C

32.8∘C

36.7∘C

38.2∘C

Q. $10 g$ of ice at $0^{\circ} C$ is mixed with $100 g$ of water at $5 0^{\circ} C$ in a calorimeter. The final temperature of the mixture is [Specific heat of water $= 1 c a l g^{- 1}^{\circ} C^{- 1}$ , latent heat of fusion of ice $= 80 c a l g^{- 1}$ ]

$31. 2^{\circ} C$

$32. 8^{\circ} C$

$36. 7^{\circ} C$

$38. 2^{\circ} C$