To raise the temperature of 5 kg of water from 20 0C to 30 0C, a 2-kg iron bar is heated and then dropped into the water. What should the temperature of the bar be (ciron = 0.46 kJ/kg 0C)

heat lost by iron + heat gained by water = 0.

[mass Fe x specific heat Fe x (Tfinal-Tinitial)] + [mass H2O x specific heat water x (Tfinal-Tinitial)] = 0
just one unknown,

To determine the temperature of the iron bar, we can use the principle of heat transfer between objects. The heat gained by the water will be equal to the heat lost by the iron bar.

Let's use the formula for heat transfer:

Q = mcΔT

Where:
Q is the heat transferred,
m is the mass of the object,
c is the specific heat capacity of the object, and
ΔT is the change in temperature.

For the water:
Q_water = m_water * c_water * ΔT_water

For the iron bar:
Q_iron = m_iron * c_iron * ΔT_iron

Since the total heat gained by the water is equal to the total heat lost by the iron bar, we have:

Q_water = Q_iron

Now let's substitute the values given:
m_water = 5 kg
c_water = 4.18 kJ/kg °C (specific heat capacity of water)
ΔT_water = (30 °C - 20 °C) = 10 °C
m_iron = 2 kg
c_iron = 0.46 kJ/kg °C (specific heat capacity of iron)
ΔT_iron = (T_iron - 20 °C) (temperature change of the iron bar)

Using the equation Q_water = Q_iron, we can solve for T_iron:

m_water * c_water * ΔT_water = m_iron * c_iron * ΔT_iron

Plugging in the values:

5 kg * 4.18 kJ/kg °C * 10 °C = 2 kg * 0.46 kJ/kg °C * (T_iron - 20 °C)

209 kJ = 0.92 kJ/°C * (T_iron - 20 °C)

Dividing both sides by 0.92 kJ/°C:

T_iron - 20 °C = 209 kJ / (0.92 kJ/°C)

T_iron - 20 °C ≈ 227 °C

Finally, adding 20 °C to both sides:

T_iron ≈ 227 °C + 20 °C

T_iron ≈ 247 °C

Therefore, the temperature of the iron bar should be approximately 247 °C.

To find the temperature of the iron bar, we can use the principle of heat transfer and the concept of conservation of energy.

The heat gained by the water is equal to the heat lost by the iron bar. We can calculate the heat gained by the water using the equation:

Qwater = mass × specific heat capacity × change in temperature

Given that the mass of the water (mwater) is 5 kg, the specific heat capacity of water (cwater) is 4.18 kJ/kg°C, and the change in temperature (ΔT) is 30°C - 20°C = 10°C, we can calculate the heat gained by the water:

Qwater = 5 kg × 4.18 kJ/kg°C × 10°C
Qwater = 209 kJ

Now, since the heat lost by the iron bar is equal to the heat gained by the water, we can calculate the heat lost by the iron bar:

Qbar = 209 kJ

The heat lost by the iron bar can be calculated using the equation:

Qbar = mass × specific heat capacity × change in temperature

Given that the mass of the iron bar (mbar) is 2 kg and the specific heat capacity of iron (ciron) is 0.46 kJ/kg°C, we can calculate the change in temperature of the iron bar:

Qbar = 2 kg × 0.46 kJ/kg°C × ΔTbar

Since the initial temperature of the iron bar is not given, we'll use the final temperature of the water as the reference temperature for the iron bar. Let's assume the initial temperature of the bar is Tinitial:

209 kJ = 2 kg × 0.46 kJ/kg°C × (30°C - Tinitial)

Simplifying the equation:

Tinitial = 30°C - (209 kJ / (2 kg × 0.46 kJ/kg°C))
Tinitial ≈ 30°C - 227.17°C
Tinitial ≈ -197.17°C

Therefore, the initial temperature of the iron bar should be approximately -197.17°C.