Question

10 g of ice at -10°C is mixed with 10 g of water at 0°C .The amount of heat required to raise the temperature of the mixture is 10°C Celsius is ??​

Answer 1

Theory ;-

Energy required to change state of a body from solid to liquid = mLᶠ

Where, m = mass of the body

L = Latent heat of fusion = 80calg⁻¹

Energy required to change temperature in same state for

• Ice = mSⁱᶜᵉΔΤ

where, m = mass of ice

Sⁱᶜᵉ = Specific heat of ice = 0.5calg⁻¹K⁻¹

ΔΤ = Change in temperature.

• Water = mSʷᵃᵗᵉʳΔΤ

where, m = mass of water

Sʷᵃᵗᵉʳ = Specific heat of water = 1calg⁻¹K⁻¹

ΔΤ= Change in temperature.

Solution :-

To raise the temperature of mixture by 10℃, you have to follow these steps :-

1. 10g of ice at (-10℃) to 10g of ice at (0℃).
2. 10g of ice at (0℃) to 10g of water at (0℃)
3. 10g of water at (0°C) to 10g of water at (10℃)
4. Initial 10g of water at (0℃) to 10g of water at (10℃).

So, in steps 1, 2, 3 and 4, energy is required

Net energy required

= Q₁ + Q₂ + Q₃ + Q₄

= (10)(0.5)(10) + (10)(80) + (10)(1)(10) + (10)(1)(10)

= 50 + 800 + 100 + 100

= 1050 calories of heat is required

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