Question

Calculate the initial temperature of liquid \[A\] of mass\[400g\] having a specific heat capacity of \[2.4Jk{g^{ - 1}}{K^{ - 1}}\]when it is mixed with liquid\[B\] of mass \[750g\] having a specific heat capacity of\[1.6Jk{g^{ - 1}}{K^{ - 1}}\] at\[40^\circ C\]. The final temperature of mixture becomes \[25^\circ C\].
A. \[15^\circ C\]
B. \[7.5^\circ C\]
C. \[4.2^\circ C\]
D. \[6.25^\circ C\]

Answer 1

Hint:As we are given in the question about heat from \[A\]and \[B\], so we know if there is not heat loss in system then total heat loss will be equal to total heat gain as in this total heat given to \[A\]will be equal to total heat taken from \[B\] . And heat will be calculated through the formula \[H = M \times C \times T\]

Complete step-by-step solution:
As in the given question we are provided with two liquids as,\[A\]and \[B\]
And we have mass of \[A\]as, \[{M_A} = 0.4kg\]
And specific heat of \[A\]as, \[{C_A} = 2.4Jk{g^{ - 1}}{K^{ - 1}}\]
And let temperature gain in A as, \[{T_A} = \left( {25 - t} \right)^\circ C\]
And we have mass of \[B\] as, \[{M_B} = 0.75kg\]
And specific heat of \[B\]as, \[{C_B} = 1.6Jk{g^{ - 1}}{K^{ - 1}}\]
And temperature loss in \[B\] will be, \[{T_B} = \left( {40 - 25} \right)^\circ C\]
\[{T_B} = 15^\circ C\]
Heat taken from \[B\] will be,
\[{H_B} = {M_B} \times {C_B} \times {T_B}\]
\[{H_B} = 0.75 \times 1.6 \times 15\]
\[{H_B} = 18J\]
Therefore it will be equal to heat taken by \[A\],
\[{H_A} = {M_A} \times {C_A} \times {T_A}\]
\[{H_A} = 0.4 \times 2.4 \times \left( {25 - t} \right)\]
\[{H_A} = 0.96 \times \left( {25 - t} \right)\]
Therefore net heat that is heat taken from \[B\]and taken by \[A\] will be equal
 \[{H_A} = {H_B}\]
\[0.96 \times \left( {25 - t} \right) = 18\]
\[6 = 0.96t\]
\[t = 6.25^\circ C\]
Therefore the correct answer is D. option.

Additional information: As above we have discussed about specific heat so specific heat is a thermodynamic property which is used as constant and it is determined as that it is value that is required to increase the temperature of body by \[1^\circ C\]for a unit mass body.

Note:- In the given question we have assumed that there is no heat loss that is total heat taken by one will be equal to total heat given to another but if there was any type of heat loss then the answer will be different.