10 g of water at 70° C is mixed with 5 g of water at 30° C . Find the temperature of the mixture in equilibrium.

Question

10 g of water at 70° C is mixed with 5 g of water at 30° C . Find the temperature of the mixture in equilibrium. (A)  33.33° C  (B)  56.67° C  (C)  77.66° C  (D)  30° C

Tardigrade Admin · Accepted Answer

Let t oC be the temperature of the mixture. From Calorimetry principle, Heat given by 10 g of water = Heat taken by 5 g of water or m1swΔ T1=m2swΔ T ∴ 10× (70-t)=5× (t-30) ∴ t=56.67 oC

Q. 10 g of water at $70^{\circ} C$ is mixed with 5 g of water at $30^{\circ} C$ . Find the temperature of the mixture in equilibrium.

$33.33^{\circ} C$

$56.67^{\circ} C$

$77.66^{\circ} C$

$30^{\circ} C$

Let $t^{o} C$ be the temperature of the mixture. From Calorimetry principle, Heat given by 10 g of water = Heat taken by 5 g of water or $m_{1} s_{w} Δ T_{1} = m_{2} s_{w} Δ T$ $∴$ $10 \times (70 - t) = 5 \times (t - 30)$ $∴$ $t = 56.67^{o} C$

Q. 10 g of water at 70∘C is mixed with 5 g of water at 30∘C . Find the temperature of the mixture in equilibrium.

33.33∘C

56.67∘C

77.66∘C

30∘C