If anyone can please help me with this problem that would begreat. I thought I k
ID: 1662875 • Letter: I
Question
If anyone can please help me with this problem that would begreat. I thought I knew what to do but I have been trying fora while and still can't get the answer.A 180 g copper bowl contains 185 g of water, both at 20.0°C. A very hot 300 gcopper cylinder is dropped into the water, causing the water toboil, with 4.80 g being converted tosteam. The final temperature of the system is 100°C. Neglectenergy transfers with the environment. (a) How much energy (in calories) istransferred to the water as heat?
(b) How much energy (in calories) is transferred to the bowl?
(c) What is the original temperature of the cylinder?
(a) How much energy (in calories) istransferred to the water as heat?
(b) How much energy (in calories) is transferred to the bowl?
(c) What is the original temperature of the cylinder?
Explanation / Answer
Given:
mass of the copper bowl = 150g
mass of the water = 220g
both the water and copper bowl are at 20.0°C
mass of the copper cylinder = 300g
Final Temperature of the system Tf = 100Celsius
The specific heat of water is 1 cal/g·C and of copperis 0.0923 cal/g·C.
The latent heat of vaporization (Lv) of water is539 Cal/g.
Amount of water converted to steam(m') = 5g
a) The question asks for heat trasnfered to water because the some of the water tosteam we have to include the heat ofvaporization(Lv)
Q = c m (Tf - Ti) +Lv* m'
=1 cal/g·C *220g *(100°C-20°C) + 539Cal/g *5g
= 20.3kcal
b) The question asks for heat transfered to bowl
Q = c m (Tf - Ti)
Q = 0.0923cal/g·C * 150g *(100 -20 )
Q = 1.11kcal
C) The question asks for the temperature of the coppercylinder in this case Q will be the total heat transferedto water and the copper bowl = 21.41kcal
Q = c*m*( Intialtemperature - final temperature)
21.41kca; = 0.0923*300g* (Initial Temperature - 100)
Solve for the InititalTemperature and we will get 873 Celsius.