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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

JIPMERJIPMER 2015Thermal Properties of Matter

Solution:

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$