Question

If the specific heat of Lead is $0.03\text{cal }{{\text{g}}^{-1}}{{\text{ }}^{0}}{{C}^{-1}}$ , thermal capacity of 500 grams of Lead is:
$\begin{align}
  & (A)5\text{cal}{{\text{ }}^{0}}{{C}^{-1}} \\
 & (B)10\text{cal}{{\text{ }}^{0}}{{C}^{-1}} \\
 & (C)15\text{cal}{{\text{ }}^{0}}{{C}^{-1}} \\
 & (D)20\text{cal}{{\text{ }}^{0}}{{C}^{-1}} \\
\end{align}$

Answer 1

Hint: The thermal capacity of a substance is the product of specific heat of a substance and its mass. To find the thermal capacity in the above problem, we have been given the specific heat of lead and the mass for which it needs to be calculated, so it’s just a problem of simple calculation.

Complete answer:
Let us first assign some terms that we are going to use in our problem.
Let the mass of Lead be denoted by $m$. Its value has been given to us in the problem as:
$\Rightarrow m=500g$
Let the specific heat capacity of lead be given by the term $S$. According to the question, the value of specific heat of lead is given as:
$\Rightarrow S=0.03\text{cal }{{\text{g}}^{-1}}{{\text{ }}^{0}}{{C}^{-1}}$
Now, let the total thermal capacity of Lead be denoted by $C$ .
Then, using the formula for thermal heat capacity of a substance, which is given by:
$\Rightarrow C=mS$ [Let the given expression be equation number (1)]
We have, the thermal heat capacity of lead as:
Putting all the required values of the respective terms in the above equation, we get:
$\begin{align}
  & \Rightarrow C=500\times 0.03cal{{\text{ }}^{0}}{{C}^{-1}} \\
 & \therefore C=15cal{{\text{ }}^{0}}{{C}^{-1}} \\
\end{align}$
Hence, the thermal heat capacity of lead comes out to be $15cal{{\text{ }}^{0}}{{C}^{-1}}$.

Hence, option (C) is the correct option for the above problem.

Note:
In simple calculations-based problems like these, one should remember the correct formula and put the values of different parameters from the question correctly. The calculations should be checked at least twice to make sure we do not concur any errors. Also, the unit of final answer should be verified once. Since, these are comparatively easy problems, one should be quick to solve these and move on to the next question to save time in an examination.