Question

What is the Cp (in J/gC) of a material if a 50 gram sample of it at 100 C is placed into 500 grams of water at

20 C and the water temp rises to 23 C and the material temp goes down to 23 C?

Answer 1

The specific heat capacity of this material is approximately
`1.6\; \rm J\cdot g^(-1) \cdot {\left(^\circ C\right)}^(-1)`.

(Assume that there is no energy loss. Assume that the specific heat capacity of water is
`4.182\; \rm J \cdot g^(-1) \cdot {\left(^\circ C\right)}^(-1)`.)

Step-by-step explanation:

Let
`c` denote the specific heat capacity of a material. Consider a sample of that material with mass
`m`. If the temperature of that objects changes by
`\Delta T`, the corresponding energy change would be
`Q =c \cdot m \cdot \Delta T`.

Start by finding the amount of energy that water has gained from the
`50\; \rm g` sample.

• Specific heat capacity of water:
  `c(\text{water}) \approx 4.182\; \rm J \cdot g^(-1) \cdot \left(^\circ C\right)^(-1)`.
• Mass of water:
  `m(\text{water}) = 500\; \rm g`.
• Temperature change:
  `\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}`.

That
`500\; \rm g` of water would have gained:

`\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) * 500\; \rm g * 3\; \rm ^\circ C \\ &\approx 6.27* 10^(3)\; \rm J \end{aligned}`.

Assume that the
`50\; \rm g` sample and the
`500\; \rm g` of water were insulated from the surroundings. The energy that water gained would be exactly the same as the energy that the
`50\; \rm g\!` sample had released through cooling.

In other words, the
`50\; \rm g` sample has released approximately
`6.27 * 10^(3)\; \rm J` of energy as it is cooled from
`100\; \rm ^\circ C` to
`23\; \rm ^\circ C`.

• Energy change:
  `Q(\text{sample}) \approx -6.27 * 10^(3)\; \rm J` (negative because this sample has lost energy through cooling.)
• Mass of sample:
  `m = 50\; \rm g`.
• Temperature change:
  `\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}`.

Rearrange the equation
`Q = c \cdot m \cdot \Delta T` to find an expression for specific heat capacity,
`c`:

`\begin{aligned}c = (Q)/(m \cdot \Delta T)\end{aligned}`.

Apply this expression to find the specific heat capacity of the sample:

`\begin{aligned}c(\text{sample}) &= \frac{Q(\text{sample})}{m(\text{sample}) \cdot \Delta T(\text{sample})} \\ &\approx (-6.27* 10^(3)\; \rm J)/(50\; \rm g * \left(-77\; \rm ^\circ C\right)) \approx 1.6\; \rm J \cdot g^(-1) \cdot {\left(^\circ C\right)}^(-1)\end{aligned}`.