The temperature of 100 g of water is to be raised from 24°C to 90°C by adding steam to it. The mass of the steam required for this purpose is

Question

The temperature of 100 g of water is to be raised from 24°C to 90°C by adding steam to it. The mass of the steam required for this purpose is (A) 10 g (B) 12 g (C) 14 g (D) 16 g

Tardigrade Admin · Accepted Answer

Let m be the mass of steam required. By principle of calorimetry 100 × 1 × (90 - 24) = m 540 + m × 1 (100 - 90) i.e., 550 m = 100 × 66 ⇒ m = 12 g

Q. The temperature of 100 g of water is to be raised from 24°C to 90°C by adding steam to it. The mass of the steam required for this purpose is

10 g

12 g

14 g

16 g

Let m be the mass of steam required.
By principle of calorimetry
$100 \times 1 \times (90 - 24) = m 540 + m \times 1 (100 - 90)$
$i . e ., 550 m = 100 \times 66 \Rightarrow m = 12 g$

Q. The temperature of 100 g of water is to be raised from 24°C to 90°C by adding steam to it. The mass of the steam required for this purpose is

10 g

12 g

14 g

16 g

Let m be the mass of steam required. By principle of calorimetry 100×1×(90−24)=m540+m×1(100−90) i.e.,550m=100×66⇒m=12g

Let m be the mass of steam required.
By principle of calorimetry
$100 \times 1 \times (90 - 24) = m 540 + m \times 1 (100 - 90)$
$i . e ., 550 m = 100 \times 66 \Rightarrow m = 12 g$