Question:
A hot liquid \(P\) of specific heat capacity \(S\) is mixed with a cold liquid \(Q\) of mass \(40\,\mathrm{g}\) and specific heat capacity \(1.5S\). If the fall in temperature of liquid \(P\) is 3 times the rise in temperature of liquid \(Q\), then the mass of liquid \(P\) is
A hot liquid \(P\) of specific heat capacity \(S\) is mixed with a cold liquid \(Q\) of mass \(40\,\mathrm{g}\) and specific heat capacity \(1.5S\). If the fall in temperature of liquid \(P\) is 3 times the rise in temperature of liquid \(Q\), then the mass of liquid \(P\) is
Show Hint
In calorimetry,
\[
\boxed{
\text{Heat lost}=\text{Heat gained}
}
\]
i.e.,
\[
m_1c_1\Delta T_1
=
m_2c_2\Delta T_2.
\]
Updated On: Jul 15, 2026
- \(40\,\mathrm{g}\)
- \(20\,\mathrm{g}\)
- \(30\,\mathrm{g}\)
- \(50\,\mathrm{g}\)
Show Solution
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The Correct Option is B
Solution and Explanation
Step 1: Apply the principle of calorimetry.
Heat lost by hot liquid \(P\) equals heat gained by cold liquid \(Q\).
\[
m_PS(\Delta T_P)
=
40(1.5S)(\Delta T_Q).
\]
Step 2: Use the given temperature relation. Given, \[ \Delta T_P = 3\Delta T_Q. \] Substituting, \[ m_PS(3\Delta T_Q) = 40(1.5S)(\Delta T_Q). \] Cancelling \(S\) and \(\Delta T_Q\), \[ 3m_P=60, \] \[ m_P=20\,\mathrm{g}. \] Hence, \[ \boxed{20\,\mathrm{g}} \] Therefore, \[ \boxed{(B)} \] is the correct answer.
Step 2: Use the given temperature relation. Given, \[ \Delta T_P = 3\Delta T_Q. \] Substituting, \[ m_PS(3\Delta T_Q) = 40(1.5S)(\Delta T_Q). \] Cancelling \(S\) and \(\Delta T_Q\), \[ 3m_P=60, \] \[ m_P=20\,\mathrm{g}. \] Hence, \[ \boxed{20\,\mathrm{g}} \] Therefore, \[ \boxed{(B)} \] is the correct answer.
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