A 100 g metal rod is heated for 5 minutes by a 175 W electric power source and, in the
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 blockbeaker-water system.
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.
a. To calculate the specific heat capacity of the metal, we can use the equation:
Q = mcΔT
Where Q is the thermal energy transferred, m is the mass of the metal rod, c is the specific heat capacity, and ΔT is the change in temperature.
Given:
- Mass of the metal rod (m) = 100 g (or 0.1 kg)
- Change in temperature (ΔT) = (40°C - 15°C) = 25°C
We can rearrange the equation to solve for c:
c = Q / (mΔT)
To find Q, we can use the equation:
Q = Pt
Where P is the power of the electric source and t is the time.
Given:
- Power (P) = 175 W
- Time (t) = 5 minutes (or 300 seconds)
Substituting the values into the equation:
Q = 175 W * 300 s
Q = 52,500 J
Now we can calculate the specific heat capacity:
c = 52,500 J / (0.1 kg * 25°C)
c = 210 J/(kg·°C)
Therefore, the specific heat capacity of the metal is 210 J/(kg·°C).
b. i. To calculate the thermal energy lost to the environment, we can use the equation:
Q_lost = m_lost * c * ΔT_lost
Where Q_lost is the thermal energy lost, m_lost is the mass of the metal block that lost heat, c is the specific heat capacity of the metal, and ΔT_lost is the change in temperature of the metal block.
Given:
- Mass of the metal block (m_lost) = 1.85 kg
- Change in temperature of the metal block (ΔT_lost) = 2% reduction in temperature
= 2% of 37°C (initial temperature of the metal block)
= 0.02 * 37°C
= 0.74°C
Substituting the values into the equation:
Q_lost = 1.85 kg * c * 0.74°C
Note: Since we have already calculated the specific heat capacity of the metal (c) as 210 J/(kg·°C), we can use this value.
Q_lost = 1.85 kg * 210 J/(kg·°C) * 0.74°C
Q_lost = 280.07 J
Therefore, the thermal energy lost to the environment is 280.07 J.
b. ii. To find the final equilibrium temperature of the block-beaker-water system, we can use the equation:
Q_gained = Q_lost
Where Q_gained is the thermal energy gained by the water and beaker system.
Given:
- Mass of the water (m_water) = 3.4 kg
- Specific heat capacity of water (c_water) = 4190 J/(kg·°C)
- Specific heat capacity of the beaker (c_beaker) = 837 J/(kg·°C)
- Initial temperature of the water (T_water_initial) = 20°C
- Initial temperature of the beaker (T_beaker_initial) = 17°C
To find the final equilibrium temperature (T_final) of the system, we need to use the equation:
Q_gained = m_water * c_water * ΔT_water + m_beaker * c_beaker * ΔT_beaker + m_lost * c * ΔT_lost
We already know:
- The thermal energy lost to the environment (Q_lost) = 280.07 J.
Rearranging the equation, we get:
Q_gained = m_water * c_water * (T_final - T_water_initial) + m_beaker * c_beaker * (T_final - T_beaker_initial) + Q_lost
Plugging in the values, we have:
280.07 J = 3.4 kg * 4190 J/(kg·°C) * (T_final - 20°C) + 837 J/(kg·°C) * (T_final - 17°C) + 1.85 kg * 210 J/(kg·°C) * 0.74°C
Simplifying the equation, we solve for T_final:
280.07 J = 14206 J + 837 J/(kg·°C) * T_final - 16779 J + 307.65 J
Combine like terms:
0 = 837 J/(kg·°C) * T_final - 203.38 J
Solving for T_final:
T_final = 203.38 J / (837 J/(kg·°C))
T_final = 0.243 °C
Therefore, the final equilibrium temperature attained by the block-beaker-water system is approximately 0.243°C.
a. To calculate the specific heat capacity of the metal, we can use the formula:
Q = mcΔT
Where:
Q = thermal energy absorbed by the metal
m = mass of the metal
c = specific heat capacity of the metal
ΔT = change in temperature
Given data:
Mass of the metal rod (m) = 100 g = 0.1 kg
Change in temperature (ΔT) = 40 ˚C - 15 ˚C = 25 ˚C
Time (t) = 5 minutes = 5 * 60 seconds = 300 seconds
Power (P) = 175 W
First, we need to calculate the thermal energy absorbed by the metal using the power formula:
P = Q/t
Q = Pt
Q = 175 W * 300 s
Q = 52,500 J
Substituting the values into the formula, we have:
Q = mcΔT
52,500 J = (0.1 kg) * c * (25 ˚C)
Simplifying the equation, we find:
c = Q / (m * ΔT)
c = 52,500 J / (0.1 kg * 25 ˚C)
c = 21,000 J/kg·˚C
Therefore, the specific heat capacity of the metal is 21,000 J/kg·˚C.
b.
i. To calculate the thermal energy lost to the environment, we can use the formula:
Q_lost = ΔT_lost * (m_block * c_block + m_water * c_water + m_beaker * c_beaker)
Given data:
Mass of the metal block (m_block) = 1.85 kg
Change in temperature lost (ΔT_lost) = 2% reduction = 2/100 * (37 ˚C - 17 ˚C) = 0.02 * 20 ˚C = 0.4 ˚C
Specific heat capacity of water (c_water) = 4190 J/kg·˚C
Specific heat capacity of the beaker (c_beaker) = 837 J/kg·˚C
Substituting the values into the formula, we have:
Q_lost = 0.4 ˚C * (1.85 kg * c_block + 3.4 kg * 4190 J/kg·˚C + m_beaker * 837 J/kg·˚C)
ii. To find the final equilibrium temperature, we assume that no thermal energy is lost in the system. Therefore, the thermal energy lost by the block is equal to the thermal energy gained by the water and the beaker.
Q_lost = Q_gained
Substituting the values into the equation, we have:
0.4 ˚C * (1.85 kg * c_block + 3.4 kg * 4190 J/kg·˚C + m_beaker * 837 J/kg·˚C) = m_water * c_water * (T_eq - 20 ˚C)
Simplifying the equation and solving for T_eq, we find:
T_eq = [0.4 ˚C * (1.85 kg * c_block + 3.4 kg * 4190 J/kg·˚C + m_beaker * 837 J/kg·˚C)] / (m_water * c_water) + 20 ˚C
Note: The value of c_block is not given, so it needs to be provided to get the final value of T_eq.