Question

You add 50.0 g of ice initially at ‒20.0 °C to 1.00 x 102 mL warm water at 67.0 °C. When all the ice melts, the water temperature is found to be somewhere above 0 °C. Calculate the final temperature of the water.

Answer 1

Answer:

$T_2=17.8\°C$

Explanation:

Hello,

In this case, we can solve this problem by noticing that the heat lost by the warm water is gained by the ice in order to melt it:

$Q_{water}=-Q_{ice}$

In such a way, the cooling of water corresponds to specific heat and the melting of ice to sensible heat and specific heat also that could be represented as follows:

$m_{water}Cp_{water}(T_2-T_{water})=-m_{ice}\Delta H_{melting,ice}-m_{ice}\Cp_{ice}(T_2-T_{ice})$

Thus, specific heat of water is 4.18 J/g°C, heat of melting is 334 J/g and specific heat of ice is 2.04 J/g°C, thus, we can compute the final temperature as shown below:

$m_{water}Cp_{water}(T_2-T_{water})+m_{ice}Cp_{ice}(T_2-T_{ice})=-m_{ice}\Delta H_{melting,ice}\\\\T_2=\frac{-m_{ice}\Delta H_{melting,ice}+m_{water}Cp_{water}T_{water}+m_{ice}Cp_{ice}T_{ice}}{m_{water}Cp_{water}+m_{ice}Cp_{ice}} \\\\T_2=\frac{-50.0*334+100*4.18*67+50.0*2.04*-20.0}{100*4.18+50.0*2.04} \\\\T_2=17.8\°C$

Best regards.