How to describe calorimetric experiments with formulas (for physicists)

We derive a formula that describes calorimetric experiments. Answer the question.

What can we write down based on our previous knowledge

If the unknown body cools, the amount of heat released during this process can be calculated as follows:

Q_{1} = c_{k} m_{k} (t_{k2}^{\circ} - t_{k1}^{\circ})

If the water gains heat when the unknown body cools, the amount of heat gained can be calculated as follows:

Q_{2} = c_{v} m_{v} (t_{v2}^{\circ} - t_{v1}^{\circ})

At the end of the experiment, the temperatures of the water and the unknown body have equalized, so that:

t_{k2}^{\circ} = t_{v2}^{\circ}

The law of conservation of energy states that the amount of heat released from the unknown body and given to the water must equal:

Q_{1} = Q_{2}

We derive the final formula for the experiment

We know that

Q_{1} = Q_{2}

Let's substitute the expressions for $Q_{1}$ and $Q_{2}$ found in the previous bubble into this expression

c_{k} m_{k} (t_{k2}^{\circ} - t_{k1}^{\circ}) = c_{v} m_{v} (t_{v2}^{\circ} - t_{v1}^{\circ})

So that $t_{k2}^{\circ} = t_{v2}^{\circ}$ , we can bring both of these temperatures under the same notation, for example $t^{\circ}$ , so we get

c_{k} m_{k} (t^{\circ} - t_{k1}^{\circ}) = c_{v} m_{v} (t^{\circ} - t_{v1}^{\circ})

By converting, we get the formula for the specific heat of the substance of the body from the unknown material

c_{k} = - \frac{c_{v} m_{v} (t^{\circ} - t_{v1}^{\circ})}{m_{k} (t^{\circ} - t_{k1}^{\circ})}

How we calculate

To calculate the specific heat of a body from an unknown material, we need

We use the formula to calculate

c_{k} = - \frac{c_{v} m_{v} (t^{\circ} - t_{v1}^{\circ})}{m_{k} (t^{\circ} - t_{k1}^{\circ})}