1 kg of ice at -20° C is mixed with 2 kg of water at 90° 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 -1 K -1 and 2.09 kJ kg -1- K -1, respectively.)

Question

1 kg of ice at -20° C is mixed with 2 kg of water at 90° 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 -1 K -1 and 2.09 kJ kg -1- K -1, respectively.) (A) 30° C (B) 0° C (C) 80° C (D) 45° C

Tardigrade Admin · Accepted Answer

Let final temperature of mixture is T° C. Then, Heat lost by 2 kg water at 90° C to cool down at T° C = Heat gained by 1 kg ice at -20° C to reach at 0° C + Heat gained by 1 kg ice at 0° C to change its state from ice to water + Water 1 kg formed at 0° C is now absorbs heat to reach temperature of T° C ⇒ mw sw Δ T=mi si(0-(-20° C )) ⇒ 2 × 4.18 ×(90-T)=1 × 2.09 × 20 +1 × 334.4+1 × 4.18 × T ⇒ 7524-376.2=3 × 4.18 × T T=30° C So, final temperature of mixture is 30° C.

$3 0^{\circ} C$

$0^{\circ} C$

$8 0^{\circ} C$

$4 5^{\circ} C$

30∘C

0∘C

80∘C

45∘C

$3 0^{\circ} C$

$0^{\circ} C$

$8 0^{\circ} C$

$4 5^{\circ} C$