A calorimeter of water equivalent 20 g contains 180 g of water at 25° C. 'm' grams of steam at 100° C is mixed in it till the temperature of the mixure is 31° C. The value of 'm' is close to (Latent heat of water =540 cal g -1, specific heat of water =1 cal g-1 . ° C -1)

Question

A calorimeter of water equivalent 20 g contains 180 g of water at 25° C. 'm' grams of steam at 100° C is mixed in it till the temperature of the mixure is 31° C. The value of 'm' is close to (Latent heat of water =540 cal g -1, specific heat of water =1 cal g-1 . ° C -1) (A) 2.6 (B) 2 (C) 4 (D) 3.2

Tardigrade Admin · Accepted Answer

( Cal /20 gm ) ( H 2 O /180 gm ) ( text Sterm / m ) 25° C 25° C 100° C 200 × 1 ×(31-25) = m × 540+ m × 1 ×(100-31)

Q. A calorimeter of water equivalent 20g contains 180g of water at 25∘C. 'm' grams of steam at 100∘C is mixed in it till the temperature of the mixure is 31∘C. The value of 'm' is close to (Latent heat of water =540calg−1, specific heat of water =1calg−1∘C−1)

2.6

2

4

3.2

20gmCal​180gmH2​O​m Sterm ​ 25∘C25∘C100∘C 200×1×(31−25) =m×540+m×1×(100−31)

$\frac{C a l}{20 g m} \frac{H _{2} O}{180 g m} \frac{Sterm}{m}$
$2 5^{\circ} C 2 5^{\circ} C 10 0^{\circ} C$
$200 \times 1 \times (31 - 25)$
$= m \times 540 + m \times 1 \times (100 - 31)$