Please give the solution and diagram of this. Give an explansion pls. A 40-g blo
ID: 1778497 • Letter: P
Question
Please give the solution and diagram of this. Give an explansion pls. A 40-g block of ice is cooled to -78 °C and is then added to 560 g of water in an 80-g copper calorimeter at a temperature of 25 °C. Determine the final temperature of the system consisting of the ice, water, and calorimeter. (If not all the ice melts, determine how much ice is left.) Remember that the ice must first warm to 0 °C, melt, and then continue warming as water. (The specific heat of copper and ice are 387 and 2090 J/kg °C, respectively. The latent heat of melting of ice is 3.33x105 J/kg.)Explanation / Answer
Cu = 0.387 J g1 K1
0.500 cal/g ·°C = 2.090 J/g°C
Heat of fusion of water = 333.55 J/g
heat to warm the ice to 0 C = (2.090 J/g°C)(75 C)(40 g) = 6270 J
heat to melt ice = (333.55 J/g)(40 g) = 13,342 J
Total heat to get ice melted = 19,612 J
combined heat capacity of liquid water and Cu calorimeter =
(560 g)(4.184 J/g°C) + (80 g)(0.387 J/gC) = 2374 J/C
After melting, liquid water + Cu have a temp. = 25 C – (19,612 J)/(2374 J/C) = 16.74 C
Now heat the "ice water" and cool the "liquid water" + Cu until T's equalize at Tf.
(2374 J/C)(16.74 -Tf) = (40 g)(4.184 J/g°C)(Tf – 0)
(2374 J/C)(16.74 C) = [(40 g)(4.184 J/g°C + 2374 J/C]Tf
Tf = .4177 C