Solveeit Logo

Question

Question: A calorimeter contains \[70.2g\]of water at \[{15.3^0}C\]. If \[143.7g\] of water at \[{36.5^0}C\] i...

A calorimeter contains 70.2g70.2gof water at 15.30C{15.3^0}C. If 143.7g143.7g of water at 36.50C{36.5^0}C is mixed with the common temperature is 28.70C{28.7^0}C. The water equivalent of the calorimeter is
(A) 15.6g15.6g
(B) 9.4g9.4g
(C) 6.3g6.3g
(D) 13.4g13.4g

Explanation

Solution

Hint:-
Use the equation heat H=mCΔTH = mC\Delta T, mm is mass of the substance, CC is the specific heat and ΔT\Delta T is the change in temperature. Water equivalent means the mass of water which can raise the same amount of temperature in this process.

Complete step by step solution:-
In this process the heat is transferred from hot to colder body.
Total heat transferred is common heat transferred in a calorimeter.
Heat is given by H=mCΔTH = mC\Delta T
mCΔT=m1CΔT1+m2CΔT2mC\Delta T = {m_1}C\Delta {T_1} + {m_2}C\Delta {T_2}
But CCis specific heat is common in all, everything is water in this problem.
Value of specific heat of water C=1calorie/gramC = 1calorie/gram
We can cancel out the specific heat.
mΔT=m1ΔT1+m2ΔT2m\Delta T = {m_1}\Delta {T_1} + {m_2}\Delta {T_2}
Calorimeter contain water m1=70.2g{m_1} = 70.2g
Temperature of calorimeter T1=15.30C{T_1} = {15.3^0}C
Amount of water added is m2=143.7g{m_2} = 143.7g
Temperature of added water T2=36.50C{T_2} = {36.5^0}C
Common temperature or the equilibrium temperature T=28.70CT = {28.7^0}C
Change in common temperature ΔT=28.70C00C=28.70C\Delta T = {28.7^0}C - {0^0}C = {28.7^0}C
Change in calorimeter temperature ΔT1=T1T=15.30C28.70C=13.40C\Delta {T_1} = {T_1} - T = {15.3^0}C - {28.7^0}C = - {13.4^0}C
Change in added water temperature ΔT2=T2T=36.50C28.70C=7.80C\Delta {T_2} = {T_2} - T = {36.5^0}C - {28.7^0}C = {7.8^0}C
mΔT=m1ΔT1+m2ΔT2m\Delta T = {m_1}\Delta {T_1} + {m_2}\Delta {T_2}
Rearrange the equation
Water equivalent of calorimeter m
m=(m1ΔT1+m2ΔT2)ΔTm = \dfrac{{({m_1}\Delta {T_1} + {m_2}\Delta {T_2})}}{{\Delta T}}
m=(70.2×(13.4))+(143×7.8)28.7m = \dfrac{{(70.2 \times ( - 13.4)) + (143 \times 7.8)}}{{28.7}}

m=1120.86940.6828.7=180.1828.7=6.27804878g   m = \dfrac{{1120.86 - 940.68}}{{28.7}} = \dfrac{{180.18}}{{28.7}} = 6.27804878g \\\ \\\

But in the option there are only two significant figures, and there only one digit after decimal point.
6.27804878g6.27804878g Approximate to 6.3g6.3g
So the answer is (C) 6.3g6.3g

Note:- In this question we don’t need the value of Specific heat of water, it cancels outs in both sides of the equation. Calorimeter is the measuring device of heat produced in any kind of reaction which produces any kind of reaction and the material properties like heat capacity, etc.