A thermometer of mass $\mathbf{0}\mathbf{.0550}\mathbf{}\mathbf{\text{kg}}$and of specific heat $\mathbf{0}\mathbf{.837}\mathbf{}\mathbf{\text{kJ/kgK}}$reads$\mathbf{15}\mathbf{.0}\mathbf{\text{°C}}$ . It is then completely immersed in of water, and it comes to the same final temperature as the water. If the thermometer then reads $\mathbf{44}\mathbf{.4}\mathbf{\text{°C}}$, what was the temperature of the water before insertion of the thermometer?

i) Mass of thermometer, $m_t=0.0550\text{kg}$

ii) Specific heat of thermometer,$c_t=0.837\frac{\text{kJ}}{\text{kg\hspace{0.17em}K}}$

iii) Initial temperature read by thermometer, $T_{ti}=15.0\text{°C}$

iv) Mass of water, $m_w=0.300\text{kg}$

v) Final temperature, $T_f=44.4\text{°C}$

vi) Specific heat of water,$c_w=4.18\frac{\text{kJ}}{\text{kg\hspace{0.17em}K}}$ .

Using the values given in the problem, and applying the concept that the heat absorbed by one object and the heat lost by the other object areequal, we can get the initial temperature of the water.

Formula:

Heat absorbed by the object via thermal radiation,$Q=Cm(T_f-T_i)$...(i)

The heat absorbed by the thermometer is equal to the heat lost by water;so, we can write the equation using equation (i) as

$c_t{m}_t(T_f-T_{ti})=c_w{m}_w(T_{wi}-T_f)$

Now solving the above equation for, we get the equation of the temperature of the water as

$\begin{array}{c}{T}_{wi}=\frac{c_t{m}_t(T_f-T_{ti})}{c_w{m}_w}+T_f\\ =\frac{\left(0.055\text{\hspace{0.17em}kg}\right)\left(0.837\frac{\text{kJ}}{\text{kg\hspace{0.17em}K}}\right)(44.4\text{\hspace{0.17em}°C}-15\text{\hspace{0.17em}°C})}{\left(4.18\frac{\text{kJ}}{\text{kg\hspace{0.17em}K}}\right)\left(0.3\text{\hspace{0.17em}kg}\right)}+\left(44.4\text{\hspace{0.17em}°C}\right)\\ =45.47\text{\hspace{0.17em}°C}\\ \approx 45.5\text{\hspace{0.17em}°C\hspace{0.17em}}\end{array}$

Hence, the value of the temperature of the water is .$45.5\text{\hspace{0.17em}°C}$