1) In this virtual lab, you will use coffee cup calorimetry to determine the spe

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 starts at 25*c and heat up to 90*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 30*c)

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

Explanation / Answer

Ans. Since addition of hot metal increase the temperature of water sample to 30.00C, the temperature at thermal equilibrium is equal to 30.00C.

# Temperature change of water, dT1 = (Thermal equilibrium – initial) temp.

= 30.00C – 26.00C = 4.00C

# Mass of water = 130.0 g

# Temperature change of metal, dT2 = 30.00C – 90.00C = -60.00C

# Mass of metal = 45.0 g

# When hot metal is placed in contact with cold water, the hot metal loses some heat, and the water gains the same amount of heat in order to attain the thermal equilibrium.

Amount of heat gained/lost by the samples is given by-

q = m s dT                            - equation 1

Where,

q = heat gained/lost

m = mass of sample

s1 = specific heat of sample

dT = Final temperature – Initial temperature

Now,

            Heat lost by metal = Heat gained by water

            Or, -q (metal) = +q (water)             ; the –ve sign indicates heat loss

            Or, -[45.0 g x s x (-60.0)0C] = 130.0 g x 4.184 J g-10C-1 (4.0)0C

            Or, 2700.0 s g 0C = 2175.68 J

            Or, s = 2175.68 J / 2700.0 g 0C = 1.241 J g-10C-1

Therefore, specific heat of the lead = 1.241 J g-10C-1

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