500 g of water and 100 g of ice at 0°C are in a calorimeter whose water equivalent is 40 g. 10 g of steam at 100° C is added to it. Then water in the calorimeter is: (Latent heat of ice = 80 cal/g, Latent heat of steam =540 cal/g)

Question

500 g of water and 100 g of ice at 0°C are in a calorimeter whose water equivalent is 40 g. 10 g of steam at 100° C is added to it. Then water in the calorimeter is: (Latent heat of ice = 80 cal/g, Latent heat of steam =540 cal/g) (A) 580 g (B) 590 g (C) 600 g (D) 610 g

Tardigrade Admin · Accepted Answer

As 1 g of steam at 100° C melts 8g of ice at 0° C. 10 g of steam will melt 8×10 g of ice at 0°C Water in calorimeter = 500 + 80 + 10g = 590g

Q. $500 g$ of water and $100 g$ of ice at $0^{°} C$ are in a calorimeter whose water equivalent is $40 g$ . $10 g$ of steam at $10 0^{\circ} C$ is added to it. Then water in the calorimeter is : (Latent heat of ice $= 80 c a l / g$ , Latent heat of steam $= 540 c a l / g$ )

$580 g$

$590 g$

$600 g$

$610 g$

As $1 g$ of steam at $10 0^{\circ} C$ melts $8 g$ of ice at $0^{\circ} C .$
$10 g$ of steam will melt $8 \times 10 g$ of ice at $0^{°} C$ Water in calorimeter $= 500 + 80 + 10 g = 590 g$

Q. 500g of water and 100g of ice at 0°C are in a calorimeter whose water equivalent is 40g. 10g of steam at 100∘C is added to it. Then water in the calorimeter is : (Latent heat of ice =80cal/g, Latent heat of steam =540cal/g)

580g

590g

600g

610g