100 g of ice at 00C is mixed with 100 g of water at 100° C. What will be the final temperature of the mixture? (Assume no heat loss)

Question

100 g of ice at 00C is mixed with 100 g of water at 100° C. What will be the final temperature of the mixture? (Assume no heat loss) (A) 20° C (B) 40° C (C) 30° C (D) 10° C

Tardigrade Admin · Accepted Answer

(m L+m s Δ θ) text ice =(m s Δ θ) text water 100(80+t)=100(100-t) ⇒ t=10° C.(m L+m s Δ θ) text ice =(m s Δ θ) text water 100(80+t)=100(100-t) ⇒ t=10° C

Q. $100 g$ of ice at $0^{0} C$ is mixed with $100 g$ of water at $10 0^{\circ} C .$ What will be the final temperature of the mixture? (Assume no heat loss)

$2 0^{\circ} C$

$4 0^{\circ} C$

$3 0^{\circ} C$

$1 0^{\circ} C$

$(m L + m s Δ θ)_{ice} = (m s Δ θ)_{water}$ $100 (80 + t) = 100 (100 - t) \Rightarrow t = 1 0^{\circ} C$ . $(m L + m s Δ θ)_{ice} = (m s Δ θ)_{water}$ $100 (80 + t) = 100 (100 - t) \Rightarrow t = 1 0^{\circ} C$

Q. 100g of ice at 00C is mixed with 100g of water at 100∘C. What will be the final temperature of the mixture? (Assume no heat loss)

20∘C

40∘C

30∘C

10∘C