Solution:
Heat gained by water = \(100 \times 1 \times (90-30) = 6000\) cal. Heat lost by steam = \(m \times 540 + m \times 1 \times (100-90) = 550m\). Equating gives \(m \approx 11\) g.
The temperature of 100 g of water is to be raised from 30 °C to 90 °C by adding steam to it. The mass of steam required to raise this temperature will be nearly (Take, \(S_W = 1\) cal/g °C, \(L = 540\) cal/g):
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