50 g ice at 0 ° C in kept in an insulating vessel and 50 g water at 100 ° C is mixed in it. Then the final temperature of the mixture is (neglect the heat loss)

Question

50 g ice at 0 ° C in kept in an insulating vessel and 50 g water at 100 ° C is mixed in it. Then the final temperature of the mixture is (neglect the heat loss) (A) 10° C (B) 0 ° C < T m < 20 ° C (C) 20 ° C (D) above 20 ° C

Tardigrade Admin · Accepted Answer

Heat taken = Heat given miLf+misw(t - 0)=mwsw(tn - t) 50× 80+50× 1× (t)=50× 1× (100 - t) 4000=50(100 - 2 t) 80=(100 - 2 t) 2t=100-80 2t=20 t=10 ° C

Q. $50 g$ ice at $0^{\circ} C$ in kept in an insulating vessel and $50 g$ water at $100^{\circ} C$ is mixed in it. Then the final temperature of the mixture is (neglect the heat loss)

$1 0^{\circ} C$

$0^{\circ} C < T m < 20^{\circ} C$

$20^{\circ} C$

above $20^{\circ} C$

Heat taken = Heat given
$m_{i} L_{f} + m_{i} s_{w} (t - 0) = m_{w} s_{w} (t_{n} - t)$
$50 \times 80 + 50 \times 1 \times (t) = 50 \times 1 \times (100 - t)$
$4000 = 50 (100 - 2 t)$
$80 = (100 - 2 t)$
$2 t = 100 - 80$
$2 t = 20$
$t = 10^{\circ} C$

Q. 50g ice at 0∘C in kept in an insulating vessel and 50g water at 100∘C is mixed in it. Then the final temperature of the mixture is (neglect the heat loss)

10∘C

0∘C<Tm<20∘C

20∘C

above 20∘C