Question

from heat class 11th​

Answer 1

The answer is 10.5
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Answer 2

Answer:

Equblirium Temprature is 6.67°C

Explanation:

Given :- 1kg of ice at - 10°C is mixed with 1 kg water at 100°C .

• we know that ,
• specific heat of water s₁ = 4.18J/(g°C ) = 1cal /(g°C)
• specific heat of ice s₂ = 0.5cal /(g°C)
• Latent heat of water L = 80cal g⁻¹
• mass of water m₁ = 1kg
• mass of ice m₂ = 1kg
• temprature of water T₁ = 100°C
• temprature of ice T₂ = -10°C

Let the final temprature be T

Here, Heat lost by 1kg water = Heat energy needed to change the temperature of ice from –10°C to 0°C + Latent heat needed to change ice at 0°C into water at 0°C + heat absorbed by water (melted ice)

⟹m₁s₁ (T₁-T) = m₂ s₂ (0- T₂ ) + m₂L + m₂s₂ (T-T₂)

⟹1×1 (100-T) = 1 × 1/2 ×10 + 1 × 80 + 1 × 1/2(T+10)

⟹100-T = 5 + 80 + T/2/ +5

⟹ 10 = T/2+T

⟹ 10 = 3T /2

T = 20/3 = 6.67°C (Approax )Answer