In lab you submerge 100 g of 41 degree Celcius iron nails in 100 g of 25 degree Celcius water (the specific heat of iron is 0.12 cal/g degrees Celcius)

A)Equate the heat gained by the water to the heat lost by the nails and find the final temperature of the water.
B)Your lab partner is surprised by the result and says that since the masses of iron and water are equal, the final water temperature should lie closer to 33 degrees Celcius, half-way between. What is your explanation?

A) Okay, let's dive into this problem! We need to equate the heat gained by the water to the heat lost by the nails. For the water, we'll use the formula Q = mcΔT, where Q is the heat gained, m is the mass of the water, c is the specific heat of water, and ΔT is the change in temperature.

So, Qwater = (100 g)(1 cal/g°C)(Tf - 25°C), where Tf is the final temperature of the water.
And for the nails, Qnails = (100 g)(0.12 cal/g°C)(41°C - Tf).

Since the heat gained by the water is equal to the heat lost by the nails, we can set up the equation:
100(1)(Tf - 25) = 100(0.12)(41 - Tf)

Now, let's shuffle things around and solve for Tf!

100(Tf - 25) = 100(0.12)(41 - Tf)

100Tf - 2500 = 12(41 - Tf)

100Tf - 2500 = 492 - 12Tf

Now, let's bring like terms together:

100Tf + 12Tf = 492 + 2500

112Tf = 2992

Tf = 2992 / 112

Tf = 26.71°C

So, the final temperature of the water is approximately 26.71 degrees Celcius.

B) Ah, the classic halfway mark confusion! The reason the final water temperature is not halfway between the initial temperatures of the water and nails is because the specific heat of iron is different from that of water. The specific heat of iron is 0.12 cal/g°C, while the specific heat of water is 1 cal/g°C.

This means that the iron nails require less heat to change their temperature compared to the water. So, when the iron nails lose heat, the water gains it at a slower rate. As a result, the final temperature of the water will be closer to the initial temperature of the nails.

So, in this case, science says no to half-baked halfway temperature theories!

A) To equate the heat gained by the water to the heat lost by the nails, we can use the principle of conservation of energy. The heat gained by the water can be calculated using the formula:

heat gained by water = mass of water × specific heat of water × change in temperature

In this case, the mass of water is 100 g, and the specific heat of water is 1 cal/g degrees Celsius (assuming it is pure water). The change in temperature is the final temperature of the water minus the initial temperature of the water. Let's call the final temperature of the water as "T" (in degrees Celsius).

heat gained by water = 100 g × 1 cal/g degrees Celsius × (T - 25 degrees Celsius)

Now, let's calculate the heat lost by the nails. The heat lost by the nails can be calculated using the same formula, but with the mass and specific heat of iron nails.

heat lost by nails = mass of nails × specific heat of iron × change in temperature

In this case, the mass of the iron nails is 100 g, and the specific heat of iron is 0.12 cal/g degrees Celsius. The change in temperature is the initial temperature of the nails minus the final temperature of the water (T).

heat lost by nails = 100 g × 0.12 cal/g degrees Celsius × (41 degrees Celsius - T)

Since the heat gained by the water is equal to the heat lost by the nails, we can set up the equation:

100 g × 1 cal/g degrees Celsius × (T - 25 degrees Celsius) = 100 g × 0.12 cal/g degrees Celsius × (41 degrees Celsius - T)

Simplifying the equation, we get:

(T - 25) = 0.12 × (41 - T)

Solving for T:

T - 25 = 4.92 - 0.12T
1.12T = 25 + 4.92
1.12T = 29.92
T ≈ 26.71

So, the final temperature of the water would be approximately 26.71 degrees Celsius.

B) The intuition that the final water temperature should lie closer to 33 degrees Celsius may be based on the equal masses of the iron and water. However, this assumption disregards the different specific heat capacities of iron and water. Since the specific heat capacity of iron is significantly lower than that of water, the change in temperature of the iron nails will have a greater impact on the overall temperature change of the system. Therefore, the final water temperature will be closer to the initial temperature of the nails rather than the midpoint between the two initial temperatures.

A) To find the final temperature of the water after the heat transfer, we need to equate the heat gained by the water to the heat lost by the nails. We can use the equation:

Q = mcΔT

Where Q is the heat transfer, m is the mass, c is the specific heat, and ΔT is the change in temperature.

For the water:
Q_water = m_water * c_water * ΔT_water

For the nails:
Q_nails = m_nails * c_iron * ΔT_nails

Since the heat lost by the nails is equal to the heat gained by the water, we can set these two equations equal to each other:

m_water * c_water * ΔT_water = m_nails * c_iron * ΔT_nails

Substituting the given values:
100g * c_water * ΔT_water = 100g * 0.12 cal/g°C * ΔT_nails

We can rearrange the equation to solve for ΔT_water:

ΔT_water = (m_nails * c_iron * ΔT_nails) / (m_water * c_water)

Substituting the values given:
ΔT_water = (100g * 0.12 cal/g°C * (41°C - ΔT_water)) / (100g * c_water)

Now we can solve for ΔT_water:

ΔT_water = (12 cal * (41°C - ΔT_water)) / c_water

ΔT_water * c_water = 492°C * c_water - 12 cal * ΔT_water

ΔT_water * c_water + 12 cal * ΔT_water = 492°C * c_water

ΔT_water * (c_water + 12 cal) = 492°C * c_water

ΔT_water = (492°C * c_water) / (c_water + 12 cal)

Now we can substitute the specific heat value for water (for example, let's say it is 1 cal/g°C):

ΔT_water = (492°C * 1 cal/g°C) / (1 cal/g°C + 12 cal)

ΔT_water = (492°C * 1 cal/g°C) / 13 cal

ΔT_water = 37.8°C

Therefore, the final temperature of the water would be 37.8°C.

B) Your lab partner's assumption that the final water temperature should lie closer to 33°C, halfway between the initial temperatures of the nails and water, is incorrect. This assumption is based on the incorrect notion that because the masses of iron and water are equal, the final temperature would be the average of the initial temperatures.

However, in reality, the specific heat capacity (c) of a substance plays a significant role in determining how much heat it can absorb or release. In this case, iron has a much lower specific heat capacity (0.12 cal/g°C) compared to water (let's assume 1 cal/g°C). This means that iron can absorb or release less heat per gram per degree Celsius than water can.

Due to the difference in specific heat capacities, the iron nails lose heat to the water more quickly than the water gains heat from the nails. As a result, the final temperature of the water will be closer to the initial temperature of the nails, rather than the average of the initial temperatures.

In this particular scenario, the final temperature of the water is determined by calculating the exact heat transfer based on the specific heat capacities and masses of both substances.

Heat gained = Heat loss

Let final temperature be x,

Specific heat of Iron:
0.12 cal/g°C = 502 J/kg °C

Specific Heat of Water: 4186 J/kg °C

Heat gained by water = mass x specific heat capacity x change in temperature
=100(4186)* (x-25)
=(418600x-10465000)J

Heat lost by nails = mass x specific heat capacity x change in temperature
=100(502)x (41-x)
=(2058200-50200x)J

Therefore:
418600x + 50200x= 2058200+10465000
468800x = 12523200
x= 27.7 °C