a 100 g iron bar at 100 C is placed in 200 g of water at 20 C. If specific heat of iron is 0.11 cal/G*c to what final temperature will the iron bar cool?

30.11

To find the final temperature to which the iron bar will cool down, we can use the principle of conservation of heat. The heat lost by the iron bar will be equal to the heat gained by the water.

The formula for heat transfer is:

Q = mcΔT

where Q is the heat transferred, m is the mass, c is the specific heat, and ΔT is the change in temperature.

Let's calculate the heat transferred by the iron bar:

Q_iron = m_iron * c_iron * ΔT_iron

Given:
m_iron = 100 g
c_iron = 0.11 cal/g°C

The change in temperature of the iron bar is:

ΔT_iron = Tf - Ti
= Tf - 100°C

Now let's calculate the heat transferred by the water:

Q_water = m_water * c_water * ΔT_water

Given:
m_water = 200 g
c_water = 1 cal/g°C (specific heat of water)

The change in temperature of the water is:

ΔT_water = Tf - Tw
= Tf - 20°C

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

Q_iron = Q_water

m_iron * c_iron * (Tf - 100°C) = m_water * c_water * (Tf - 20°C)

Let's plug in the values and solve for Tf:
(100 g) * (0.11 cal/g°C) * (Tf - 100°C) = (200 g) * (1 cal/g°C) * (Tf - 20°C)

11(Tf - 100) = 200(Tf - 20)
11Tf - 1100 = 200Tf - 4000
189Tf = 2900
Tf ≈ 15.34°C

Therefore, the iron bar will cool down to approximately 15.34°C.

To determine the final temperature to which the iron bar will cool, we need to calculate the heat transferred from the iron bar to the water using the formula:

Q = m * c * ΔT

Where:
Q is the heat transferred (in calories),
m is the mass of the substance (in grams),
c is the specific heat capacity (in cal/g°C), and
ΔT is the change in temperature (in °C).

First, let's calculate the heat transferred from the iron bar to the water:

Q_iron = m_iron * c_iron * ΔT_iron

Q_water = m_water * c_water * ΔT_water

Since the heat lost by the iron bar is equal to the heat gained by the water (assuming no heat is lost to the surroundings):

Q_iron = -Q_water

Let's substitute the given values and variables:

m_iron = 100 g
c_iron = 0.11 cal/g°C
ΔT_iron = Tf - 100°C (final temperature of the iron bar measured from its initial temperature)
m_water = 200 g
c_water = 1 cal/g°C (specific heat capacity of water)
ΔT_water = Tf - 20°C (final temperature of the water measured from its initial temperature)

Now, we can set up the equation:

m_iron * c_iron * (Tf - 100) = -m_water * c_water * (Tf - 20)

Substituting the given values:

100g * 0.11 cal/g°C * (Tf - 100) = -200g * 1 cal/g°C * (Tf - 20)

11(Tf - 100) = -200(Tf - 20)

11Tf - 1100 = -200Tf + 4000

11Tf + 200Tf = 1100 + 4000

211Tf = 5100

Tf = 5100 / 211

Tf ≈ 24.17°C

Therefore, the iron bar will cool to approximately 24.17°C as its final temperature.