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A mass m1 of a substance of specific heat capacity c1 at temperature t1 is mixed with a mass m2 of other substance of specific heat capacity c2 at a lower temperature t2. Deduce the expression for the

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Question

A mass m1 of a substance of specific heat capacity c1 at temperature t1 is mixed with a mass m2 of other substance of specific heat capacity c2 at a lower temperature t2. Deduce the expression for the temperature t of the mixture. State the assumption made, if any.

Numerical
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Solution

A mass m1 of a substance A of specific heat capacity c1 at temperature T1 is mixed with a mass m2 of other substance B of specific heat capacity c2 at a lower temperature T2 and final temperature of the mixture becomes T.

Fall in temperature of substance A = T1 – T

Rise in temperature of substance B = T – T2

Heat energy lost by A = m1 × c1 × fall in temperature = m1c1 (T1 – T)

Heat energy gained by B = m2× c2 × rise in temperature = m2c2 (T – T2)

If no energy lost in the surrounding, then by the principle of mixtures,

Heat energy lost by A = Heat energy gained by B

m1C1 (T1 - T) = m2C2 (T - T2)

or m1 C1 T1 - m1 C1 T = m2 C2 T - m2 C2 T2

∴ [m2 C2 T2 + m1 C1 T1] = m2 C2 T + m1 C1 T

[m1 C1 + m2 C2] T = [m1 C1 T1 + m2 C2 T2]

T = `(m_1 C_1 T_1 + m_2  C_2 T_2)/(m_1 C_1 + m_2 C_2)`

Here, we have assumed that there is no loss of heat energy.

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Chapter 11: Calorimetry - EXERCISE-11(A) [Page 270]

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Selina Concise Physics [English] Class 10 ICSE
Chapter 11 Calorimetry
EXERCISE-11(A) | Q 22. | Page 270

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