temperature and heat?

Question

2kg water at room temperature 20 celcius is placed inside a refrigerator with an internal temperature of 2 celcius.
a. calculate the amount of heat Qc that must be removed to cool the water to 2 celcius.
b. the coefficient of performance?
c.how much work done?
d. what is the total heat dumped by the refrigerator into the kitchen

Answer 1

a) multiply the specific heat of water (1 cal/g degree or something like that) by the change in temp and the mass to get the heat removed.

b) The coefficient of perf for an IDEAL fridge is Tc / (Th-Tc) where Th and Tc are the hot and cold temperatures (measured in Kelvin, so add 273.15 to celsius).

c) Divide a by b

d) The heat dumped into the kitchen is c plus a.

Here's a nice website that derives the equations if you want to understand it at more than the plug-n-chug level that I described.

edit--the answer below is wrong--c is not equal to a of course (converted or no). The fridge isn't perfectly efficient. Since they don't give you the efficiency, assume that it has ideal efficiency--ie use the equation I gave for (b)

Answer 2

a. By definition 1 cal is the heat needed to rise/cool 1 g of water temperature by 1 oC. Since you have 2000 g to cool by 18 oC your extracted heat Qc = 2000*18 = 36000 cal = 36 kcal.

b. coefficient of performance = Qc/Qt
where Qt = heat dumped into a kitchen.

c. Since the heat is equivalent to work 1 kcal = 4186 J the work done = 150696 J = 150696 Ws = 41.86 Wh = 0.04186 kWh = 15372 kpm.

d. heat dumped into a kitchen = Qc/coefficient of performance.

Since you don't know Qt nor coefficient of performance of your fridge you can't calculate b. nor d.