Question

How much heat is lost from a block of solid gold when 2000 g of this gold (Au) cools from $1064{}^\circ \text{C}$ to $25{}^\circ \text{C}$? [${{\text{C}}_{\text{p}}}(\text{Au)}=0.128$]

Answer 1

Hint: The amount of heat transferred when the temperature of an object changes can be calculated by the given formula:
\[\text{q}=\text{m}{{\text{C}}_{\text{p}}}\text{dT}\]
Here $q$ is the heat transferred when the mass of a substance is subjected to change in temperature dT and ${{\text{C}}_{\text{p}}}$ is the specific heat capacity of that substance.

Complete step-by-step answer:
The heat capacity (C) of a substance is the amount of heat (q) it absorbs or releases when it undergoes a temperature change$\left( \Delta \text{T} \right)$of $1{}^\circ \text{C}$.
\[\text{C}=\dfrac{\text{q}}{\Delta \text{T}}\]
The unit of heat capacity is $\text{J }{}^\circ {{\text{C}}^{-1}}$.
The specific heat capacity ${{\text{C}}_{\text{p}}}$ of a substance is the amount of heat required to raise the temperature of 1 gram of a substance by $1{}^\circ \text{C}$.
\[{{\text{C}}_{\text{p}}}=\dfrac{\text{q}}{\text{m}\Delta \text{T}}\]
The SI unit of specific heat capacity is $\text{J }{}^\circ {{\text{C}}^{-1}}{{\text{g}}^{-1}}$.
If we know the amount of a substance and its specific heat capacity, we can determine the amount of heat gained or lost by the substance by measuring the temperature change.
The specific heat capacity and mass of the substance is always a positive number, but $\Delta \text{T}$ and q can be either positive or negative.
-If a substance gains energy, its final temperature becomes higher than its initial temperature. Then, $\Delta \text{T}>\text{0}$ and the value of q will be positive.
- If a substance loses energy, its final temperature becomes lower than its initial temperature. Then $\Delta \text{T}<\text{0}$ and the value of q will be negative.
In the question, we are given the following values:
$\begin{align}
  & \text{m}=2000\text{ g} \\
 & \Delta \text{T}={{\text{T}}_{\text{final}}}-{{\text{T}}_{\text{initial}}}=25{}^\circ \text{C}-1064{}^\circ \text{C}=-1039{}^\circ \text{C} \\
 & {{\text{C}}_{\text{p}}}\left( \text{Au} \right)=0.128\text{ J}{}^\circ {{\text{C}}^{-1}}{{\text{g}}^{-1}} \\
\end{align}$
Substituting these values in the equation of specific heat capacity, we get:
$\begin{align}
  & \text{0}\text{.128 J }{}^\circ {{\text{C}}^{-1}}{{\text{g}}^{-1}}=\dfrac{\text{q}}{2000\text{ g}\times \left( -1039 \right){}^\circ \text{C}} \\
 & \Rightarrow \text{q}=-265984\text{ J} \\
 & \Rightarrow \text{q}=-265.984\text{ kJ} \\
\end{align}$
Hence, 265.98 kJ of heat is lost from the block of solid gold.

Note: The negative sign of q represents that the heat is lost from the substance to the surrounding and the positive sign of q represents that the heat is gained by the substance from the surrounding. The heat capacity is an extensive property but the specific heat capacity is an intensive property that means it does not depend upon the amount of matter.