To solve this question, we need to calculate two things in sequence:
1. The amount of heat energy required to raise the temperature of 10 kg of iron from 20°C to 150°C.
2. The increase in temperature of 10 kg of water when the same amount of heat energy is supplied.
### Step 1: Calculating the heat energy required for iron
First, let's identify the given data for iron:
- Mass of iron ([tex]\(m_{iron}\)[/tex]) = 10 kg
- Initial temperature of iron ([tex]\(T_{initial,iron}\)[/tex]) = 20°C
- Final temperature of iron ([tex]\(T_{final,iron}\)[/tex]) = 150°C
- Specific heat capacity of iron ([tex]\(c_{iron}\)[/tex]) = 460 J/kg°C
We use the formula for heat energy:
[tex]\[ Q = m \times c \times \Delta T \][/tex]
Where:
- [tex]\( Q \)[/tex] is the heat energy
- [tex]\( m \)[/tex] is the mass
- [tex]\( c \)[/tex] is the specific heat capacity
- [tex]\( \Delta T \)[/tex] is the change in temperature ([tex]\( T_{final} - T_{initial} \)[/tex])
Now, let's calculate the change in temperature for iron:
[tex]\[ \Delta T_{iron} = T_{final,iron} - T_{initial,iron} \][/tex]
[tex]\[ \Delta T_{iron} = 150°C - 20°C = 130°C \][/tex]
Now, we can calculate the heat energy:
[tex]\[ Q_{iron} = m_{iron} \times c_{iron} \times \Delta T_{iron} \][/tex]
[tex]\[ Q_{iron} = 10 \, \text{kg} \times 460 \, \text{J/kg°C} \times 130 \, \text{°C} \][/tex]
[tex]\[ Q_{iron} = 10 \times 460 \times 130 \][/tex]
[tex]\[ Q_{iron} = 598,000 \, \text{J} \][/tex]
So, the amount of heat energy required to raise the temperature of 10 kg of iron from 20°C to 150°C is [tex]\( 598,000 \)[/tex] Joules.
### Step 2: Calculating the increase in temperature for water
Now that we know the amount of heat energy supplied ([tex]\( Q_{iron} = 598,000 \, \text{J} \)[/tex]), we need to find the increase in temperature for 10 kg of water when the same energy is supplied.
Let's identify the given data for water:
- Mass of water ([tex]\(m_{water}\)[/tex]) = 10 kg
- Specific heat capacity of water ([tex]\(c_{water}\)[/tex]) = 4200 J/kg°C
Again, we use the formula for heat energy, rearranged to solve for the change in temperature ([tex]\( \Delta T \)[/tex]):
[tex]\[ Q = m \times c \times \Delta T \][/tex]
[tex]\[ \Delta T_{water} = \frac{Q}{m_{water} \times c_{water}} \][/tex]
Now, let's calculate the increase in temperature for water:
[tex]\[ \Delta T_{water} = \frac{Q_{iron}}{m_{water} \times c_{water}} \][/tex]
[tex]\[ \Delta T_{water} = \frac{598,000 \, \text{J}}{10 \, \text{kg} \times 4200 \, \text{J/kg°C}} \][/tex]
[tex]\[ \Delta T_{water} = \frac{598,000}{42,000} \][/tex]
[tex]\[ \Delta T_{water} \approx 14.238 \, \text{°C} \][/tex]
So, if the same amount of heat energy (598,000 J) is supplied to 10 kg of water, the increase in temperature of the water will be approximately [tex]\( 14.23 \)[/tex]°C.
Therefore, the answers are:
- The amount of heat energy required to raise the temperature of 10 kg of iron from 20°C to 150°C is [tex]\( 598,000 \)[/tex]J.
- If the same amount of heat is supplied to 10 kg of water, the increase in temperature of the water will be [tex]\( 14.23 \)[/tex]°C.
