coffee cup calorimetry

Question

1) In this virtual lab, you will use coffee cup calorimetry to determine the specific heat, c, of a metal. 

Imagine that, in lab, you record the mass of a piece of metal, mmetal, as 

m(metal) =45.00 g 

In the next step of the lab (as shown in the animation to the left) you heat up the piece of metal. Play the animation, and record the metal\'s highest temperature, Tmetal. 

Tmetal= _____c* 

(it start at 22*c and heat up to 85*c) 

2)In another step of this lab, you record the mass of water (mwater) in a coffee cup as 

m(water)= 130.00 g 

What is the initial temperature of the water, Twater1, in the coffee cup as shown to the left? 

Twater 1= ____c* 

In the next step of the lab (as shown in the animation to the left) you place the hot piece of metal into the cool water. Play the animation and record the final temperature of the water, Twater2. 

Twater 2= ____c* 

(it start at 26*c and heat up to 29*c) 

3) With the information above and the specific heat of water, 4.184 J/(g·°C), you calculate the specific heat, c, of the metal.

Answer 1

Hello Trinity

First, we need to calculate the amount of heat absorbed by the water when the metal is added to the calorimeter.

q = m c (delta T)

**I am unable to type the symbol for delta as Wyzant has for some reason removed the ability for us tutors to type them.

mass of water = 130.00 g

c = 4.184 J / g oC

Initial Temp, (Ti) = 26 oC

Final Temp. (Tf) = 29 oC

q absorbed by water = 130.00 x 4.184 x (Tf - Ti)

= 130.00 x 4.184 x (29 - 26)

= + 1631.8 J

According to the Law of Conservation of Energy, heat absorbed by water = heat lost by the metal. The value for q will be the same, however, since heat was lost by the metal, the sign for q will be negative (-).

q lost by the metal = - 1631.8 J

mass of metal = 45.00 g

Ti of metal = 85 oC

Tf of metal = 29 oC

Metal was initially heated to 85 oC before adding it to the water in the calorimeter. Water had an initial temperature of 26 oC and gained heat from the metal resulting in an increase in temp. to 29 oC. The metal lost heat resulting in a decrease in its temp. from 85 to 29 oC.

specific heat of the metal, c = q / m (delta T)

= - 1631.8 / 45.00 x (29 - 85)

= - 1631.8 / - 2520

= 0.648 J / g oC