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*10^5\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

Step-by-step 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}*2090\text{ J/kg C}*(0-(-73)+0.025\text{ kg }*3.33*10^5\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)

Thus, the final temperature of the system is

`15.72\text{ C}`