91Ó°ÊÓ

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:

$m\mathrm{Δ}{C}_p\mathrm{Δ}T$

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

$T_{i,\text{water}}=\frac{\left(m_{p,\text{lhermo}}\mathrm{Δ}{C}_{p,\text{siass}}+m_{\text{water}}\mathrm{Δ}{C}_{p,\text{water}}\right)\mathrm{Δ}{T}_f-m_{\text{thermo}}\mathrm{Δ}{C}_{\text{p,siass}}\mathrm{Δ}{T}_{i,\text{thermo}}}{m_{\text{water}}\mathrm{Δ}{C}_{p,\text{water}}}$

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

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

And there's the glass's specific heat:

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

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

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

The initial and final temperatures in Kelvin:

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

Then the substituting values are:

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

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

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.