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Standard X

Physics

Method of Mixtures

1 Kg of ice a...

Question

1 Kg of ice at -20 $^{\circ}$ C is mixed with 2 Kg of water at 90 $^{\circ}$ C. Assuming that there is no loss of energy to the environment, what will be the final temperature of the mixture? (Assume latent heat of ice = 334.4 KJ/Kg, specific heat of water and ice are 4.18 kJ/(kg.K) and 2.09 kJ/(kg.K), respectively.)

30 $^{\circ}$ C

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0 $^{\circ}$ C

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80 $^{\circ}$ C

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45 $^{\circ}$ C

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Solution

The correct option is A
30 $^{\circ}$ C

$m_{i} s_{i} (Δ T) + m_{i} L + m_{i} . s_{w} (T - 0) = m_{w} s_{w} (90 - T)$
$1 \times 2.09 (20) + 1 \times 334.4 + 1 \times 4.18 \times T = 2 \times 4.18 \times (90 - T)$
T = 60 - 30 = 30 $^{\circ}$ C

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The correct option is A 30∘C misi(ΔT)+miL+mi.sw(T−0)=mwsw(90−T) 1×2.09(20)+1×334.4+1×4.18×T=2×4.18×(90−T) T = 60 - 30 = 30∘ C

The correct option is A
30 $^{\circ}$ C

$m_{i} s_{i} (Δ T) + m_{i} L + m_{i} . s_{w} (T - 0) = m_{w} s_{w} (90 - T)$
$1 \times 2.09 (20) + 1 \times 334.4 + 1 \times 4.18 \times T = 2 \times 4.18 \times (90 - T)$
T = 60 - 30 = 30 $^{\circ}$ C