A piece of iron of mass 0.2 kg is kept inside a furnace, till it attains the temperature of the furnace. The hot piece of iron is then dropped into a calorimeter containing 0.24 kg of water at 20° C. The mixture attains an equilibrium temperature of 60°C. The temperature of the furnace isa)500°Cb)400°Cc)506.80 °Cd)406.80 °CCorrect answer is option 'C'. Can you explain this answer?

Question

A piece of iron of mass 0.2 kg is kept inside a furnace, till it attains the temperature of the furnace. The hot piece of iron is then dropped into a calorimeter containing 0.24 kg of water at 20&deg; C. The mixture attains an equilibrium temperature of 60&deg;C. The temperature of the furnace isa)500&deg;Cb)400&deg;Cc)506.80 &deg;Cd)406.80 &deg;CCorrect answer is option 'C'. Can you explain this answer?

Answer

Use heat lost = heat gain . but the specific heat of iton must be given in the ques

Answer

°C. The final temperature of the mixture is found to be 30 °C. Determine the specific heat capacity of iron. Assume no heat is lost to the surroundings.

To solve this problem, we can use the principle of conservation of energy. The heat lost by the iron is equal to the heat gained by the water:

Heat lost by iron = Heat gained by water

The heat lost or gained can be calculated using the formula:

q = mcΔT

where q is the heat gained or lost, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.

Let's calculate the heat lost by the iron:

q_iron = m_iron * c_iron * ΔT_iron

Given:
m_iron = 0.2 kg
ΔT_iron = (final temperature - initial temperature) = 30 - initial temperature

Now, let's calculate the heat gained by the water:

q_water = m_water * c_water * ΔT_water

Given:
m_water = 0.24 kg
c_water = specific heat capacity of water = 4,186 J/kg°C (approximately)
ΔT_water = (final temperature - initial temperature) = 30 - 20 = 10°C

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

q_iron = q_water

m_iron * c_iron * ΔT_iron = m_water * c_water * ΔT_water

0.2 kg * c_iron * (30 - initial temperature) = 0.24 kg * 4,186 J/kg°C * 10°C

Now, we can solve for c_iron:

c_iron = (0.24 kg * 4,186 J/kg°C * 10°C) / (0.2 kg * (30 - initial temperature))

Let's assume the initial temperature of the iron is T_iron_initial. Substituting the values:

c_iron = (0.24 kg * 4,186 J/kg°C * 10°C) / (0.2 kg * (30 - T_iron_initial))

Now, we have all the values to calculate the specific heat capacity of iron.

Answer

I guess b

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