Let's tackle the first problem involving the cube of iron and the ice block. This is a classic calorimetry question where we need to find the initial temperature of the iron cube based on the heat transfer to the ice. We'll break it down step by step.
Understanding the Problem
We have a cube of iron with the following properties:
- Density of iron: 8000 kg/m³
- Specific heat capacity of iron: 470 J/kg/K
The cube is heated to a high temperature and then placed on a large block of ice at 273 K (0°C). The ice melts due to the heat transferred from the iron cube, and we need to find the initial temperature of the cube.
Key Concepts
When the iron cube is placed on the ice, it loses heat, which is used to melt the ice. The heat lost by the iron cube will equal the heat gained by the ice. The heat gained by the ice can be calculated using the formula:
Q = m × L_f
Where:
- Q: Heat gained by the ice (in joules)
- m: Mass of ice melted (in kg)
- L_f: Latent heat of fusion of ice (336,000 J/kg)
Calculating the Mass of Ice Melted
To find the mass of ice melted, we need to know the volume of the iron cube. Let's denote the volume of the cube as V. The mass of the iron cube can be calculated as:
m_iron = density_iron × V
Now, since the cube sinks into the water, the volume of the cube will equal the volume of the water displaced, which is the mass of the melted ice divided by the density of ice:
m_ice = V × density_ice
Setting Up the Equation
Using the heat lost by the iron cube:
Q_iron = m_iron × c_iron × (T_initial - T_final)
Where:
- T_initial: Initial temperature of the iron cube
- T_final: Final temperature (273 K)
Now, we can equate the heat lost by the iron to the heat gained by the ice:
m_iron × c_iron × (T_initial - 273) = m_ice × L_f
Substituting Values
We know:
- c_iron = 470 J/kg/K
- L_f = 336,000 J/kg
- density_ice = 900 kg/m³
Now, substituting the mass of the ice melted:
m_ice = (density_ice × V)
And for the iron cube:
m_iron = (density_iron × V)
Now we can simplify our equation:
(density_iron × V) × c_iron × (T_initial - 273) = (density_ice × V) × L_f
The volume V cancels out:
density_iron × c_iron × (T_initial - 273) = density_ice × L_f
Final Calculation
Substituting the known values:
8000 kg/m³ × 470 J/kg/K × (T_initial - 273) = 900 kg/m³ × 336,000 J/kg
Calculating the right side:
900 × 336,000 = 302,400,000 J
Now, substituting back:
8000 × 470 × (T_initial - 273) = 302,400,000
Calculating the left side:
3,760,000 (T_initial - 273) = 302,400,000
Now, divide both sides by 3,760,000:
T_initial - 273 = 80.2
Finally, adding 273 to both sides gives:
T_initial = 353.2 K
So, the initial temperature of the iron cube is approximately 353 K, which matches the answer you provided.
Moving to the Second Problem
Now, let’s look at the second problem involving the brick dropped into the river. We need to calculate the thermal energy generated when the brick hits the water.
Understanding the Energy Conversion
The brick has a mass of 4 kg and is dropped from a height of 2 m into the river. The potential energy (PE) of the brick at the height can be calculated using the formula:
PE = m × g × h
Where:
- m: Mass of the brick (4 kg)
- g: Acceleration due to gravity (approximately 9.81 m/s²)
- h: Height (2 m)
Calculating Potential Energy
Substituting the values:
PE = 4 kg × 9.81 m/s² × 2 m = 78.48 J
Finding Thermal Energy
Since 80% of the potential energy is converted into thermal energy, we can calculate the thermal energy (TE) as follows:
TE = 0.8 × PE
Substituting the potential energy we calculated:
TE = 0.8 × 78.48 J = 62.784 J
Converting Joules to Calories
To convert joules to calories, we use the conversion
