Question

Exercise 5: I was changing 34 around too much that I just made it a separate exercise. Suppose 310. grams of ethanol (ethyl alcohol) is in an aluminum cup of 90.0 grams. Both of these are at 30.0C. A mass m of ice at - 8.5C is taken from a freezer and added to the alcohol in the cup. The final temperature of all the components is 18.0C. Assuming no heat was lost from the system, calculate the mass m of the ice added. (ans: somewhat less than 25g) .], EM! Ex 6: Repeat the half-sheet exercise of Tuesday to calculate the final temperature of the mixture with these variations: A) there is 84,300 Joules lost to the environment that is, following the method from class, you bring the water down to 22.0C, SPILL 84,300J, and put the remainder back into the system. Remember: 600kg of wax starts at 22.0C and 2.00kg of water starts at 98.0C. This solid wax has a spec. heat of 1250J/kgK, a melting point of 55.0C, and a liquid specific heat of 1800. J/kgK. B Assume is 1.50kg of wax and 1.50kg of water, but all other parameters stay the same initial temps, T of melting, specific heats, heat of fusion.) For either of these, if the final T is 55.0C, how much of the wax remains solid? Exercise 7: Stinkywinkium has a specific heat of 12303/kgk in liquid form, and 933J/kgk in solid form. It melts at 45.0C (L = 225,000J/kg) and boils at a high temp we don't care about, but when it boils, it sure is stinky. Suppose you have 600 grams of stinky winkium at 95.0C and add it to 301 grams of ice at 0.0C. Use the method from class to find the final state of the mixture (temperature of water stinkywinkium and whether the latter is solid, liquid, or partly melted) ASSUMING the heat exchange is only between the stinky winkium and the water. Answer: 45.0C, less than 70 grams melted - you find out how much.) B) Show that if the system loses 2.20E4J to the environment, the final temperature is very close to 41C.)

Answer 1

Specific heat of ethanol Se = 2.46 x me = 310g T = 30° C Specific heat of Aluminium Sa = 0.9 mg = 90g T; = 30° C Specific heat of ice and water S; = 2.108 x Sw = 4.186 X J/gC m = mg T; = -8.5°C latent heat of ice L = 333.55 x final temperature Tj = 18°C from principle of calorimetry Heat loss by ethanol and aluminium cup will be equal to heat gain by ice to melt and then to reach up to final temp 18°C 0; = Q + Now Heat loss by ethanol and aluminium cup to reach up to final temp 18°C le + MS, 4T, + M, S, 4T 310 x 2.46 X (30 - 18) + 90 x 0.9 X (30 – 18) = 10123.2 10123.2J

heat gain by ice to reach up to final temp 18°C m; S; AT; + mL + Mw Sw 4Tw (m) x (2.108) X (8.5) + (m) x (333.55) + m x 4.186 X (18) 426.816m NOW Heat loss = heat gain 10123.2 = 426.816m m= - 10123.2 426.816 m= 23.7g