A 100 g metal rod is heated for 5 minutes by a 175 W electric power source and, in the

Question

process, its temperature changes from 15˚C to 40˚C.
a. Calculate the specific heat capacity of this metal assuming that there are no losses
in thermal energy during this process.
b. A bigger block of this same metal with a mass of 1.85 kg at 37˚C is
transported from the front of a laboratory to the back where it is placed,
together with 3.4 kg of water (c = 4190 J.kg-1
. ˚C -1
) at 20 ˚C , in an
insulated beaker (c =837 J.kg-1
.˚C-1
) at 17˚C. During the transportation,
the block loses thermal energy to the surrounding leading to a 2%
reduction in its temperature by the time it is placed in the beaker.
i. Calculate the thermal energy that was lost to the environment.
ii. Find the final equilibrium temperature that is attained by the block￾beaker-water system.

Answer 1

a. First, we need to calculate the amount of thermal energy absorbed by the metal rod:

Q = Pt = (175 W)(5 min x 60 s/min) = 52500 J

Next, we can use the formula for specific heat capacity:

Q = mcΔT

Solving for c, we get:

c = Q / (mΔT) = 52500 J / (0.1 kg x 25 ˚C) = 2100 J/kg.˚C

Therefore, the specific heat capacity of the metal is 2100 J/kg.˚C.

b. i. We can use the formula for thermal energy:

Q = mcΔT

to calculate the thermal energy lost by the metal block. We know the initial temperature of the block (37˚C) and the final temperature after losing 2% of its temperature (0.98 x 37 = 36.26˚C), so we can calculate the temperature change (ΔT = 0.74 ˚C).

Using the specific heat capacity calculated in part (a) and the mass of the metal block (1.85 kg), we get:

Q = mcΔT = (1.85 kg)(2100 J/kg.˚C)(0.74 ˚C) = 2843.5 J

Therefore, the thermal energy lost by the metal block during transportation is 2843.5 J.

ii. To find the final equilibrium temperature of the system, we need to use the formula for heat transfer:

Q = mcΔT

where Q is the thermal energy transferred, m is the mass of the substance, c is its specific heat capacity, and ΔT is the temperature change.

We can apply this formula separately for the metal block, the water, and the beaker, then set them equal to each other because the total thermal energy in the system is conserved:

Qblock + Qwater + Qbeaker = 0

The initial temperature of the block is 37˚C, and its final temperature after losing thermal energy during transportation is 36.26˚C. The final temperature of the water and beaker will be the same, so let's call it T.

For the metal block, we have:

Qblock = mcΔT = (1.85 kg)(2100 J/kg.˚C)(-0.74 ˚C) = -2843.5 J

For the water, we have:

Qwater = mcΔT = (3.4 kg)(4190 J/kg.˚C)(T - 20 ˚C)

For the beaker, we have:

Qbeaker = mcΔT = (0.1 kg)(837 J/kg.˚C)(T - 17 ˚C)

Plugging these values into the conservation of thermal energy equation gives:

-2843.5 J + (3.4 kg)(4190 J/kg.˚C)(T - 20 ˚C) + (0.1 kg)(837 J/kg.˚C)(T - 17 ˚C) = 0

Simplifying and solving for T, we get:

T = 23.36˚C

Therefore, the final equilibrium temperature of the system is 23.36˚C.