Question

A 50.0 g thermometer is used to measure the temperature of

200 mL of water. The specific heat of the thermometer, which

is mostly glass, is 750 J/kg K, and it reads 20.0C while lying

on the table. After being completely immersed in the water, the

thermometer’s reading stabilizes at 71.2C. What was the actual

water temperature before it was measured?

Short answer

$T_{i,\text{water}}=73.50\left[{}^{\circ }\mathrm{C}\right]$

Step-by-step solution

Given information

The thermometer has a mass of $50.0$a specific heat of $75J/kgK,$and an initial temperature of $20.0^0$. The final temperature is $71.2°\mathrm{C}.$We have of $200\mathrm{mL}$water.

The equation to calculate the energy of each element is:

$m\Delta C_p\Delta T$

Step :2 Calculation

Solving for $T_{i,\text{water}}$:

$T_{i,\text{water}}=\frac{\left(m_{p,\text{lhermo}}\Delta C_{p,\text{siass}}+m_{\text{water}}\Delta C_{p,\text{water}}\right)\Delta T_f-m_{\text{thermo}}\Delta C_{\text{p,siass}}\Delta T_{i,\text{thermo}}}{m_{\text{water}}\Delta C_{p,\text{water}}}$

Now, we need to know how much the mass$200\mathrm{mL}$of water is:

$200[\mathrm{mL}]\Delta \frac{1\left|g\right|}{1[\mathrm{mL}]}=200[\mathrm{g}]$

And there's the glass's specific heat:

$$\begin{gathered} C_{p_{\text{slass}}}=750\left[\frac{J}{kg\Delta K}\right] \\ =0.750\left[\frac{J}{g\Delta K}\right] \end{gathered}$$

The following information can be found in a table of specific heat capacity:

$C_{p,\text{water}}=4.181\left[\frac{\mathrm{J}}{g\Delta K}\right]$

The initial and final temperatures in Kelvin:

$$\begin{gathered} T_{i,\text{thermo}}=20\left[{}^{\circ }\mathrm{C}\right]+273.15=293.15[\mathrm{K}] \\ {T}_f=71.2\left[{}^{\circ }\mathrm{C}\right]+273.15=344.35\left[K\right] \end{gathered}$$

Then the substituting values are:

$T_{i,\text{water}}=\frac{\left(50.0\left[g\mid \Delta 0.750\left[\frac{J}{{}_g\Delta K}\right]+200\left[g\mid \Delta 4.181\left[\frac{J}{g4K}\right]\right)\Delta 344.35\left[K\right]-50.0\left[g\mid \Delta 0.750\left[\frac{J}{{}_g\Delta K}\right]\Delta 293.15\left[K\right]\right.\right.\right.}{200\left|g\right|\Delta 4.181\left[\frac{J}{g\Delta K}\right]}$

$T_{i,\text{water}}=346.65\left[K\right]=73.50\left[{}^{\circ }\mathrm{C}\right]$

Even little items have the ability to provide or take energy and alter measurements. When great precision is required, situations like these must be considered.