One kilogram of ice at 0° C is mixed with one kilogram of water at 80° C. The final temperature of the mixture is (Take specific heat of water =4200 kJ / kg - ° C, Latent heat of ice =336 kJ / kg )

Question

One kilogram of ice at 0° C is mixed with one kilogram of water at 80° C. The final temperature of the mixture is (Take specific heat of water =4200 kJ / kg - ° C, Latent heat of ice =336 kJ / kg ) (A) 0° C (B) 40° C (C) 50°C (D) 60°C

Tardigrade Admin · Accepted Answer

It is known that For water and ice mixing θmix=(mW θW-(mi Li/CW)/mi+mW) θ operatornamemix=(mW θW-(mi Li/cW)/mi+mW) ∵ m i = m w ⇒ θmix=(θ W -( L i / CW )/2) =(80-(336/4.2)/2)=0° C

Q. One kilogram of ice at $0^{\circ} C$ is mixed with one kilogram of water at $8 0^{\circ} C$ . The final temperature of the mixture is (Take specific heat of water $= 4200 k J / k g -^{\circ} C$ , Latent heat of ice $= 336 k J / k g$ )

$0^{\circ} C$

$4 0^{\circ} C$

$5 0^{\circ} C$

$6 0^{\circ} C$

Q. One kilogram of ice at 0∘C is mixed with one kilogram of water at 80∘C. The final temperature of the mixture is (Take specific heat of water =4200kJ/kg−∘C, Latent heat of ice =336kJ/kg )

0∘C

40∘C

50∘C

60∘C

It is known that For water and ice mixing θmix​=mi​+mW​mW​θW​−CW​mi​Li​​​ θmix​=mi​+mW​mW​θW​−cW​mi​Li​​​ ∵mi​=mw​ ⇒θmix​=2θW​−CWLi​​​ =280−4.2336​​=0∘C

Q. One kilogram of ice at $0^{\circ} C$ is mixed with one kilogram of water at $8 0^{\circ} C$ . The final temperature of the mixture is (Take specific heat of water $= 4200 k J / k g -^{\circ} C$ , Latent heat of ice $= 336 k J / k g$ )

$0^{\circ} C$

$4 0^{\circ} C$

$5 0^{\circ} C$

$6 0^{\circ} C$