19 g of water at 30° C and 5 g of ice at - 20° C are mixed together in a calorimeter. What is the final temperature of the mixture? Given specific heat of ice = 0.5 cal g-1 °C -1 and latent heat of fusion of ice = 80 cal g-1

Question

19 g of water at 30° C and 5 g of ice at - 20° C are mixed together in a calorimeter. What is the final temperature of the mixture? Given specific heat of ice = 0.5 cal g-1 °C -1 and latent heat of fusion of ice = 80 cal g-1 (A) 0° C (B) - 5° C (C) 5° C (D) 10° C

Tardigrade Admin · Accepted Answer

Let the final temperature of mixture be

t^{\circ} C

.
Heat lost by water in calories,

H_{1} = 19 \times 1 \times (30 - t)

= 570 - 19 t

Heat taken by ice,

H_{2} = m s_{p} Δ t + m L + m s_{w} t

= 5 \times (0.5) 20 + 5 \times 80 + 5 \times 1 \times t

According to principle of calorimetry,

H_{1} = H_{2}

5 \times (0.5) \times 20 + 5 \times 80 + 5 t = 570 - 19 t

\Rightarrow 24 t = 570 - 450 = 120

\Rightarrow t = 5^{\circ} C

Q. 19g of water at 30°C and 5g of ice at −20°C are mixed together in a calorimeter. What is the final temperature of the mixture? Given specific heat of ice = 0.5calg−1°C−1 and latent heat of fusion of ice = 80calg−1

0° C

- 5° C

5° C

10° C

Let the final temperature of mixture be t∘C. Heat lost by water in calories, H1​=19×1×(30−t) =570−19t Heat taken by ice, H2​=msp​Δt+mL+msw​t =5×(0.5)20+5×80+5×1×t According to principle of calorimetry, H1​=H2​ 5×(0.5)×20+5×80+5t=570−19t ⇒24t=570−450=120 ⇒t=5∘C

Q. $19 g$ of water at $30° C$ and $5 g$ of ice at $- 20° C$ are mixed together in a calorimeter. What is the final temperature of the mixture? Given specific heat of ice = $0.5 c a l g^{- 1} ° C^{- 1}$ and latent heat of fusion of ice = $80 c a l g^{- 1}$