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Question: An unknown metal of mass 192g heated to a temperature of \({100}^{\circ}C\) was immersed into a bras...

An unknown metal of mass 192g heated to a temperature of 100C{100}^{\circ}C was immersed into a brass calorimeter of mass 128g containing 240g of water at a temperature of 8.4C{8.4}^{\circ}C . Calculate the specific heat of the unknown metal if water temperature stabilizes at 21.5C{21.5}^{\circ}C. (Specific heat of brass is 394Jkg1K1394Jk{{g}^{-1}}{{K}^{-1}})
(A)1232Jkg1K1 (B)458Jkg1K1 (C)654Jkg1K1 (D)916Jkg1K1 \begin{aligned} & (A)1232Jk{{g}^{-1}}{{K}^{-1}} \\\ & (B)458Jk{{g}^{-1}}{{K}^{-1}} \\\ & (C)654Jk{{g}^{-1}}{{K}^{-1}} \\\ & (D)916Jk{{g}^{-1}}{{K}^{-1}} \\\ \end{aligned}

Explanation

Solution

A brass calorimeter is used to measure the specific heat of any metal with the help of known data on specific heat of water and brass. It uses the principle that the energy released by the unknown metal is the total energy absorbed by the water-brass system. At equilibrium, all the three components in a brass calorimeter exist at the same temperature. We shall use this to solve our question.

Complete step-by-step solution:
Let the specific heat of the unknown metal be given by S. And the specific heat of water and brass be given by SW{{S}_{W}} and SB{{S}_{B}} respectively. Then, it has been given to us that:
SW=4180Jkg1K1 SB=394Jkg1K1 \begin{aligned} & \Rightarrow {{S}_{W}}=4180Jk{{g}^{-1}}{{K}^{-1}} \\\ & \Rightarrow {{S}_{B}}=394Jk{{g}^{-1}}{{K}^{-1}} \\\ \end{aligned}
Now, the formula for the energy lost or gained by any substance without undergoing change of state is given by:
Q=mS(T)\Rightarrow Q=mS\left( \vartriangle T \right)
Where,
m is the mass of substance.
S is the specific heat. And,
T\vartriangle T is the change in temperature.
Thus, using the above equation, we can write the amount of energy lost by the unknown metal is equal to:
Q=192×S×(10021.5)\Rightarrow Q=192\times S\times (100-21.5) [Let this expression be equation number (1)]
Now, the energy gained by the brass calorimeter is equal to:
Q=128×394(21.58.4)+240×4180×(21.58.4)\Rightarrow {{Q}^{'}}=128\times 394(21.5-8.4)+240\times 4180\times (21.5-8.4)
[Let this above expression be equation number (1)]
Since, energy lost by the metal is equal to the energy gained by the calorimeter, therefore equation number (1) and (2) must be equal. Therefore, on equating them. We get:
192×S×(10021.5)=128×394(21.58.4)+240×4180×(21.58.4)\Rightarrow 192\times S\times (100-21.5)=128\times 394(21.5-8.4)+240\times 4180\times (21.5-8.4)
15072S=13824038.2Jkg1K1 S=917.2Jkg1K1 S916Jkg1K1 \begin{aligned} & \Rightarrow 15072S=13824038.2Jk{{g}^{-1}}{{K}^{-1}} \\\ & \Rightarrow S=917.2Jk{{g}^{-1}}{{K}^{-1}} \\\ & \therefore S\approx 916Jk{{g}^{-1}}{{K}^{-1}} \\\ \end{aligned}
Hence, the specific heat of the metal comes out to be 916Jkg1K1916Jk{{g}^{-1}}{{K}^{-1}}.
Hence, option (D) is the correct option.

Note: We should be aware of these physical processes. In calorimetry, it is important to note that, not only the water but the brass container too absorbs heat. Also, in lengthy calculations like one performed above, one should be careful and check his/her calculations at every step of the procedure.