Question

proposed solution scenario: building a berm
disko bay ilulissat icefjord
if a concrete berm were built, represented by the added rectangle on top of the fjord sill, the flow of warm water would be impeded. it would be built with 2.4 x 10^14 grams of concrete which has a specific heat of 0.88 j/gc.
assume that every day, 7.04 x 10^16 grams of salt water will contact the berm wall. the water would begin with an average temperature yesterday of 4.2°c, but then after coming into contact with the wall, would drop to 3.5°c. salt water has a different specific heat capacity than pure water, at 3.85 j/(g c).

1. what would be energy absorbed by the concrete wall, assuming that it starts at - 3.2°c?

in = 2.4×10^14
c = 0.88 j/gc
δt = 3.5°c - 4.2°c
δt=-0.7

1. as the wall transfers that energy to the other side to the cooler water, what would be the final temperature of the fresh water, which starts at 2.6°c? (use your calculated mass from number 2)

Explanation:

Step1: Identify the formula for heat - energy

The formula for heat energy is $q = mc\Delta T$, where $m$ is the mass, $c$ is the specific heat capacity, and $\Delta T$ is the change in temperature.

Step2: Determine the values for the concrete - wall problem

Given $m = 2.4\times10^{14}$ g, $c = 0.88$ J/(g°C), and $\Delta T$ needs to be calculated from the temperature of the water and the initial temperature of the concrete. The heat lost by the salt - water is equal to the heat gained by the concrete. The change in temperature of the salt - water is $\Delta T_{water}=3.5 - 4.2=- 0.7$°C. The heat lost by the salt - water is $q_{water}=m_{water}c_{water}\Delta T_{water}$, where $m_{water}=7.04\times10^{16}$ g and $c_{water}=3.85$ J/(g°C). So $q_{water}=7.04\times10^{16}\times3.85\times(- 0.7)$ J. This heat is gained by the concrete.
For the concrete, $q_{concrete}=m_{concrete}c_{concrete}\Delta T_{concrete}$. We know $m_{concrete}=2.4\times10^{14}$ g and $c_{concrete}=0.88$ J/(g°C). Since $q_{concrete}=q_{water}$, we can solve for $\Delta T_{concrete}$.
First, calculate $q_{water}=7.04\times10^{16}\times3.85\times(-0.7)=7.04\times10^{16}\times(- 2.695)=-1.89788\times10^{17}$ J.
Then, from $q_{concrete}=m_{concrete}c_{concrete}\Delta T_{concrete}$, we have $\Delta T_{concrete}=\frac{q_{concrete}}{m_{concrete}c_{concrete}}$. Substituting $q_{concrete}=1.89788\times10^{17}$ J, $m_{concrete}=2.4\times10^{14}$ g and $c_{concrete}=0.88$ J/(g°C), we get $\Delta T_{concrete}=\frac{1.89788\times10^{17}}{2.4\times10^{14}\times0.88}=\frac{1.89788\times10^{17}}{2.112\times10^{14}} = 898.62$°C.
The initial temperature of the concrete $T_0=-3.2$°C, so the final temperature of the concrete $T = T_0+\Delta T_{concrete}=-3.2 + 898.62=895.42$°C.

For question 4, assume the heat transferred from the concrete to the fresh - water is $q$. Let the mass of the fresh - water be $m_{fresh}$ (not given in the problem - statement completely, but we know from the heat - transfer principle $q = m_{fresh}c_{fresh}\Delta T_{fresh}$). Let the initial temperature of the fresh - water $T_{0,fresh}=2.6$°C and $c_{fresh}$ (specific heat of fresh - water, assume $c_{fresh}=4.18$ J/(g°C)).
Since the heat transferred from the concrete is $q = 1.89788\times10^{17}$ J, from $q = m_{fresh}c_{fresh}\Delta T_{fresh}$, we have $\Delta T_{fresh}=\frac{q}{m_{fresh}c_{fresh}}$. Then the final temperature of the fresh - water $T_{f,fresh}=T_{0,fresh}+\Delta T_{fresh}=2.6+\frac{1.89788\times10^{17}}{m_{fresh}\times4.18}$.

Since the mass of the fresh - water from number 2 is not given, we assume for the sake of showing the method. If we assume $m_{fresh}=M$ (a placeholder for the actual mass), then $\Delta T_{fresh}=\frac{1.89788\times10^{17}}{M\times4.18}$ and $T_{f,fresh}=2.6+\frac{1.89788\times10^{17}}{M\times4.18}$.

If we assume the heat transfer is 100% efficient and no heat is lost to the surroundings:
Let's first solve question 3 accurately.
The heat lost by the salt - water is $q = m_{water}c_{water}\Delta T_{water}$, where $m_{water}=7.04\times10^{16}$ g, $c_{water}=3.85$ J/(g°C) and $\Delta T_{water}=3.5 - 4.2=-0.7$°C.
$q=7.04\times10^{16}\times3.85\times(- 0.7)=-1.89788\times10^{17}$ J.
This is the heat gained by the concrete.
For the concrete, $q = m_{concrete}c_{concrete}\Delta T_{concrete}$, $m_{concrete}=2.4\times10^{14}$ g, $c_{concrete}=0.88$ J/(g°C)
$\Delta T_{concrete}=\frac{q}{m_{concrete}c_{concrete}}=\frac{1.89788\times10^{17}}{2.4\times10^{14}\times0.88}=\frac{1.89788\times10^{17}}{2.112\times10^{14}}=898.62$°C
Initial temperature of concrete $T_1=-3.2$°C
Final temperature of concrete $T_2=-3.2 + 898.62 = 895.42$°C

For question 4, let the mass of fresh - water be $m$.
The heat transferred $q = 1.89788\times10^{17}$ J, $c = 4.18$ J/(g°C), initial temperature $T_0 = 2.6$°C
$\Delta T=\frac{q}{mc}$
Final temperature $T=T_0+\frac{q}{mc}$

Answer:

1. The energy absorbed by the concrete wall is $1.89788\times10^{17}$ J and its final temperature is 895.42°C.
2. Without the mass of the fresh - water (from number 2), the final temperature of the fresh - water is $T = 2.6+\frac{1.89788\times10^{17}}{m\times4.18}$°C, where $m$ is the mass of the fresh - water.