Answer:
The specific heat capacity of this material is approximately [tex]1.6\; \rm J\cdot g^{-1} \cdot {\left(^\circ C\right)}^{-1}[/tex].
(Assume that there is no energy loss. Assume that the specific heat capacity of water is [tex]4.182\; \rm J \cdot g^{-1} \cdot {\left(^\circ C\right)}^{-1}[/tex].)
Explanation:
Let [tex]c[/tex] denote the specific heat capacity of a material. Consider a sample of that material with mass [tex]m[/tex]. If the temperature of that objects changes by [tex]\Delta T[/tex], the corresponding energy change would be [tex]Q =c \cdot m \cdot \Delta T[/tex].
Start by finding the amount of energy that water has gained from the [tex]50\; \rm g[/tex] sample.
- Specific heat capacity of water: [tex]c(\text{water}) \approx 4.182\; \rm J \cdot g^{-1} \cdot \left(^\circ C\right)^{-1}[/tex].
- Mass of water: [tex]m(\text{water}) = 500\; \rm g[/tex].
- Temperature change: [tex]\begin{aligned}\Delta T(\text{water}) &= T(\text{final}) - T(\text{initial}) \\ &= 23\; \rm ^\circ C - 20\; \rm ^\circ C = 3\; \rm ^\circ C\end{aligned}[/tex].
That [tex]500\; \rm g[/tex] of water would have gained:
[tex]\begin{aligned} & Q(\text{water}) \\ &= c(\text{water}) \cdot m(\text{water}) \cdot \Delta T (\text{water}) \\ &\approx 4.182\; \rm J \cdot g^{-1} \cdot \left(^\circ C\right)^{-1} \times 500\; \rm g \times 3\; \rm ^\circ C \\ &\approx 6.27\times 10^{3}\; \rm J \end{aligned}[/tex].
Assume that the [tex]50\; \rm g[/tex] sample and the [tex]500\; \rm g[/tex] of water were insulated from the surroundings. The energy that water gained would be exactly the same as the energy that the [tex]50\; \rm g\![/tex] sample had released through cooling.
In other words, the [tex]50\; \rm g[/tex] sample has released approximately [tex]6.27 \times 10^{3}\; \rm J[/tex] of energy as it is cooled from [tex]100\; \rm ^\circ C[/tex] to [tex]23\; \rm ^\circ C[/tex].
- Energy change: [tex]Q(\text{sample}) \approx -6.27 \times 10^{3}\; \rm J[/tex] (negative because this sample has lost energy through cooling.)
- Mass of sample: [tex]m = 50\; \rm g[/tex].
- Temperature change: [tex]\begin{aligned}\Delta T(\text{sample}) &= T(\text{final}) - T(\text{initial}) \\ &= 23\; \rm ^\circ C - 100\; \rm ^\circ C = -77\; \rm ^\circ C\end{aligned}[/tex].
Rearrange the equation [tex]Q = c \cdot m \cdot \Delta T[/tex] to find an expression for specific heat capacity, [tex]c[/tex]:
[tex]\begin{aligned}c = \frac{Q}{m \cdot \Delta T}\end{aligned}[/tex].
Apply this expression to find the specific heat capacity of the sample:
[tex]\begin{aligned}c(\text{sample}) &= \frac{Q(\text{sample})}{m(\text{sample}) \cdot \Delta T(\text{sample})} \\ &\approx \frac{-6.27\times 10^{3}\; \rm J}{50\; \rm g \times \left(-77\; \rm ^\circ C\right)} \approx 1.6\; \rm J \cdot g^{-1} \cdot {\left(^\circ C\right)}^{-1}\end{aligned}[/tex].
