100 g of ice (latent heat 80 cal g -1, at 0° C ) is mixed with 100 g of water (specific heat 1 cal g -1 ° C -1 ) at 80° C. The final temperature of the mixture will be

Question

100 g of ice (latent heat 80 cal g -1, at 0° C ) is mixed with 100 g of water (specific heat 1 cal g -1 ° C -1 ) at 80° C. The final temperature of the mixture will be (A) 0° C (B) 40° C (C) 80° C (D) <0° C

Tardigrade Admin · Accepted Answer

100 g ice 0° C +100 g water 80° C Q text required =m L if Δ T=80 [Qf. becomes .=0°] =100 × 80 Q text available =m C Δ T =8000 cal =100 × 1 × 80=8000 cal [Required to melt whole ice] Since Q text required =Q text avallable ∴ Whole system will be water at equillibrium and temperature would be 0° C

Q. $100 g$ of ice (latent heat $80 c a l g^{- 1}$ , at $0^{\circ} C$ ) is mixed with $100 g$ of water (specific heat $1 c a l g^{- 1}^{\circ} C^{- 1}$ ) at $8 0^{\circ} C$ . The final temperature of the mixture will be

$0^{\circ} C$

$4 0^{\circ} C$

$8 0^{\circ} C$

$< 0^{\circ} C$

Q. 100g of ice (latent heat 80calg−1, at 0∘C ) is mixed with 100g of water (specific heat 1calg−1∘C−1 ) at 80∘C. The final temperature of the mixture will be

0∘C

40∘C

80∘C

<0∘C

100g ice 0∘C+100g water 80∘C Qrequired ​=mL if ΔT=80 [Qf​ becomes =0∘] =100×80 Qavailable ​=mCΔT =8000cal =100×1×80=8000cal [Required to melt whole ice] Since Qrequired ​=Qavallable ​ ∴ Whole system will be water at equillibrium and temperature would be 0∘C

Q. $100 g$ of ice (latent heat $80 c a l g^{- 1}$ , at $0^{\circ} C$ ) is mixed with $100 g$ of water (specific heat $1 c a l g^{- 1}^{\circ} C^{- 1}$ ) at $8 0^{\circ} C$ . The final temperature of the mixture will be

$0^{\circ} C$

$4 0^{\circ} C$

$8 0^{\circ} C$

$< 0^{\circ} C$