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Question: Calculate the initial temperature of liquid \[A\] of mass\[400g\] having a specific heat capacity of...

Calculate the initial temperature of liquid AA of mass400g400g having a specific heat capacity of 2.4Jkg1K12.4Jk{g^{ - 1}}{K^{ - 1}}when it is mixed with liquidBB of mass 750g750g having a specific heat capacity of1.6Jkg1K11.6Jk{g^{ - 1}}{K^{ - 1}} at40C40^\circ C. The final temperature of mixture becomes 25C25^\circ C.
A. 15C15^\circ C
B. 7.5C7.5^\circ C
C. 4.2C4.2^\circ C
D. 6.25C6.25^\circ C

Explanation

Solution

As we are given in the question about heat from AAand BB, 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 AAwill be equal to total heat taken from BB . And heat will be calculated through the formula H=M×C×TH = M \times C \times T

Complete step-by-step solution:
As in the given question we are provided with two liquids as,AAand BB
And we have mass of AAas, MA=0.4kg{M_A} = 0.4kg
And specific heat of AAas, CA=2.4Jkg1K1{C_A} = 2.4Jk{g^{ - 1}}{K^{ - 1}}
And let temperature gain in A as, TA=(25t)C{T_A} = \left( {25 - t} \right)^\circ C
And we have mass of BB as, MB=0.75kg{M_B} = 0.75kg
And specific heat of BBas, CB=1.6Jkg1K1{C_B} = 1.6Jk{g^{ - 1}}{K^{ - 1}}
And temperature loss in BB will be, TB=(4025)C{T_B} = \left( {40 - 25} \right)^\circ C
TB=15C{T_B} = 15^\circ C
Heat taken from BB will be,
HB=MB×CB×TB{H_B} = {M_B} \times {C_B} \times {T_B}
HB=0.75×1.6×15{H_B} = 0.75 \times 1.6 \times 15
HB=18J{H_B} = 18J
Therefore it will be equal to heat taken by AA,
HA=MA×CA×TA{H_A} = {M_A} \times {C_A} \times {T_A}
HA=0.4×2.4×(25t){H_A} = 0.4 \times 2.4 \times \left( {25 - t} \right)
HA=0.96×(25t){H_A} = 0.96 \times \left( {25 - t} \right)
Therefore net heat that is heat taken from BBand taken by AA will be equal
HA=HB{H_A} = {H_B}
0.96×(25t)=180.96 \times \left( {25 - t} \right) = 18
6=0.96t6 = 0.96t
t=6.25Ct = 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 1C1^\circ Cfor 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.