Question

what mass of steam initially at 130°C is needed to warm 200 g of water in a 100- g glass container from 20.0C to 50.0 C ?​

Answer 1

Answer:

Mass of the steam required to raise the temperature of water is

$m = 2.92 \times 10^{-3} g$

Explanation:

Heat given by the steam = heat absorbed by glass + heat absorbed by water

So we have

$m_1s_1(\Delta T_1) + m_1L + m_1s_1\Delta T_2 = (m_2s_2 + m_3 s_1)(\Delta T)$

now we have

$m(1)(130 - 100) + m(2260000) + m(1)(100 - 50) = (100(0.20) + 200(1))(50 - 20)$

$m(2260080) = 220 \times 30$

$m = 2.92 \times 10^{-3} g$

#Learn

Topic : Calorimetry

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