By adding some amount of ice- the temperature of 500 g of water at 50- C- is lowered to 0- C- If the specific heat capacity of water is 4-2 J g -1 K -1 and specific specific latent heat of ice is 336 J g -1- then find the mass of ice added to it-A- 312-5 gB- 312-5 KgC- 3-125 gD- 3-125 kg

The correct option is A 312.5g Given Mass of water = 500g Initial temperature = 50^∘C , Final temperature = 0^∘C Fall in temperature = Δt = (50 0) = 50^∘C Heat ...

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Standard X

Physics

Specific Heat Capacity

By adding som...

Question

By adding some amount of ice, the temperature of 500g of water at $50^{\circ} C$ , is lowered to $0^{\circ} C$ . If the specific heat capacity of water is 4.2 J g⁻¹ K⁻¹and specific specific latent heat of ice is 336 J g⁻¹, then find the mass of ice added to it.

312.5g

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312.5Kg

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3.125g

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3.125kg

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Solution

The correct option is A
312.5g

Given Mass of water = 500g
Initial temperature = $50^{\circ} C$ , Final temperature = $0^{\circ} C$
Fall in temperature = Δt = (50-0) = $50^{\circ} C$
Heat lost by water = mcΔt
= (500kg)(4.2 J g⁻¹ K⁻¹)(50K)
= 1.05× 10⁶J
if m'g of ice is added, heat gained by it to melt to $0^{\circ} C$
=m'L = m'×336
By principle of method of mixture, Heat lost by water = Heat gained by ice
1.05× 10⁶= m'×336

$m^{'} = \frac{1.05 \times 106}{336}$ =312.5g

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, specific heat capacity of water =

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, specific latent heat of ice =

336 J g^{- 1}

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By adding some amount of ice, the temperature of 500g of water at 50∘C , is lowered to 0∘C. If the specific heat capacity of water is 4.2 J g−1 K−1 and specific specific latent heat of ice is 336 J g−1, then find the mass of ice added to it.

By adding some amount of ice, the temperature of 500g of water at $50^{\circ} C$ , is lowered to $0^{\circ} C$ . If the specific heat capacity of water is 4.2 J g⁻¹ K⁻¹and specific specific latent heat of ice is 336 J g⁻¹, then find the mass of ice added to it.