Multiple choice

A liquid of specific heat $0.3 cal/g^{o}C$ at $90^{o}C$ is mixed with another liquid of specific heat $0.5 cal/g^{o}C$ at $15^{o}C$. If the resultant temperature of mixture is $60^{o}C$ then the ratio of their masses is:

  1. 5 : 2

  2. 2 : 5

  3. 1 : 1

  4. 3 : 4
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A Correct answer
Explanation

Heat lost = Heat gained. m1 * c1 * (T1 - T) = m2 * c2 * (T - T2). m1 * 0.3 * (90 - 60) = m2 * 0.5 * (60 - 15). m1 * 0.3 * 30 = m2 * 0.5 * 45. 9 * m1 = 22.5 * m2. m1/m2 = 22.5 / 9 = 2.5 = 5/2.

AI explanation

Using the principle of heat exchange, the heat lost by the hotter liquid equals the heat gained by the colder liquid, which is mathematically written as (m1)(0.3)(90 - 60) = (m2)(0.5)(60 - 15). This simplifies to 9(m1) = 22.5(m2). The ratio of their masses m1 : m2 is 22.5 : 9, which simplifies to 5 : 2.

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