Question

A 25 g block of ice is cooled to -73 degrees celsius. It is added to 517 g of water in an 74 g copper calorimeter at a temperature of 22 degrees celsius. Find the final temperature. The specific heat of copper is 387 J/kg oC and of ice is 2090 J/kg oC. The latent heat of fusion of water is 3.33 x 10^ 5 J/kg and its specific heat is 4186 J/kg oC. Answer in units of degrees celsius.

Answer 1

Given:

the mass of the ice is

$\begin{gathered} m_1=25\text{ g} \\ m_1=0.025\text{ kg} \end{gathered}$

the temperature of the ice is

$T_1=-73\text{ C}$

The specific heat of the ice is

$k_1=2090\text{ K/kg C}$

latent heat of the ice is

$L=3.33\times10^5\text{ J/kg}$

The mass of the water is

$\begin{gathered} m_2=517\text{ g} \\ m_2=0.517\text{ kg} \end{gathered}$

The temperature of the water is

$T_2=22\text{ C}$

Specific heat of the water is

$k_2=4186\text{ J/kg C}$

The mass of the copper is

$\begin{gathered} m_3=74\text{ g} \\ m_3=0.074\text{ kg} \end{gathered}$

The temperature of the copper is

$T_3=22\text{ C}$

and specific heat of the copper is

$k_3=387\text{ J/kg C}$

Required: the final temperature

Explanation:

to solve this problem we will use the concept of calorimetry

that is given as

$\Delta Q=0.....(1)$

froheat required to melt the ice form -73 to 0 is given by

$Q_1=m_1k_1(0-T_1)+m_1L$

plugging all the values in the above relation we get

$\begin{gathered} Q_1=0.025\text{ kg}\times2090\text{ J/kg C}\times(0-(-73)+0.025\text{ kg }\times3.33\times10^5\text{ J/kg} \\ Q_1=12139.25\text{ J} \end{gathered}$

now calculate for whole system

heat required to increase the temperature of melted ice is

$Q_2=m_1k_2(T-0)$

heat required water to increase its temperature

$Q_3=m_2k_2(T-T_2)$

heat required to increase the temperature of the copper is

$Q_4=m_3k_3(T-T_3)$

now from the equation (1)

$\begin{gathered} Q_1+Q_2+Q_3+Q_3=0 \\ 12139.25\text{ J +0.025}\times4186(T-0)+0.517\text{ kg}\times4186\text{ J/kg}(T-22)+0.074\text{ kg}\times387\text{ J/kg C }((T-22)=0 \\ T=\frac{36102}{2297.29} \\ T=15.72\text{ C} \end{gathered}$

Thus, the final temperature of the system is

$15.72\text{ C}$