Question

A 825 g iron block is heated to 352 degrees C and is placed in an insulated container (of negligible heat capacity) containing 40.0 g of water at 20.0 degrees C. What is the final equilibrium temperature of the water and iron? specific heat of water = 4186 J/(kg K), specific heat of water vapor = 2090 J/(kg K), specific heat of iron= 560 J/(kg K), latent heat of vaporization for water= 2.26*10^6 J/kg

Answer 1

To calculate the final equilibrium temperature of the water and iron, use the equation Q_water + Q_iron = 0. Plug in the mass, specific heat capacity, and initial and final temperatures for water and iron, and solve the equation to find the final equilibrium temperature.

Step-by-step explanation:

To calculate the final equilibrium temperature of the water and iron, we can use the equation:

Q_water + Q_iron = 0

Where Q_water is the heat absorbed by the water and Q_iron is the heat absorbed by the iron.

First, let's calculate Q_water:

Q_water = m_water * c_water * ΔT

Where m_water is the mass of water, c_water is the specific heat capacity of water, and ΔT is the change in temperature. Plugging in the values:

Q_water = 40.0g * 4.186J/(g°C) * (T_final - 20.0°C)

Next, let's calculate Q_iron:

Q_iron = m_iron * c_iron * ΔT

Where m_iron is the mass of iron, c_iron is the specific heat capacity of iron, and ΔT is the change in temperature. Plugging in the values:

Q_iron = 825g * 560J/(g°C) * (T_final - 352°C)

Since Q_water + Q_iron = 0, we have:

m_water * c_water * (T_final - 20.0°C) + m_iron * c_iron * (T_final - 352°C) = 0

Plugging in the values and solving for T_final:

40.0g * 4.186J/(g°C) * (T_final - 20.0°C) + 825g * 560J/(g°C) * (T_final - 352°C) = 0

Now, you can solve the equation to find the final equilibrium temperature.

Answer 2

Answer : The final equilibrium temperature of the water and iron is, 537.12 K

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

`q_1=-q_2`

`m_1* c_1* (T_f-T_1)=-m_2* c_2* (T_f-T_2)`

where,

`c_1` = specific heat of iron = 560 J/(kg.K)

`c_1` = specific heat of water = 4186 J/(kg.K)

`m_1` = mass of iron = 825 g

`m_2` = mass of water = 40 g

`T_f` = final temperature of water and iron = ?

`T_1` = initial temperature of iron =
`352^oC=273+352=625K`

`T_2` = initial temperature of water =
`20^oC=273+20=293K`

Now put all the given values in the above formula, we get:

`(825* 10^(-3)kg)* 560J/(kg.K)* (T_f-625K)=-(40* 10^(-3)kg)* 4186J/(kg.K)* (T_f-293K)`

`T_f=537.12K`

Therefore, the final equilibrium temperature of the water and iron is, 537.12 K