A hot 0.25-kg piece of iron is placed in a calorimeter containing room temperature water. How much heat is lost by the iron as it cools from 125 degree C to 75 degree C? I must have the given/work and answer also how much heat is gained by the water
WHATS THE ANSWER
To calculate the heat lost by the iron as it cools from 125°C to 75°C, you can use the equation:
Q = mcΔT
Where:
Q = heat lost/gained
m = mass of the object (in this case, the iron)
c = specific heat capacity of the material (iron)
ΔT = change in temperature
Given:
Mass of iron, m = 0.25 kg
Temperature change, ΔT = (75°C - 125°C) = -50°C (negative because the iron is cooling)
First, we need to calculate the specific heat capacity of iron (c). Using a reference, we can find that the specific heat capacity of iron is around 450 J/kg°C.
Therefore:
c = 450 J/kg°C
Now we can calculate the heat lost by the iron:
Q = mcΔT
Q = (0.25 kg)(450 J/kg°C)(-50°C)
Q = -5625 J
The negative sign indicates that heat is lost.
Therefore, the iron loses 5625 Joules of heat as it cools from 125°C to 75°C.
To calculate the heat gained by the water in the calorimeter, we use the same equation:
Q = mcΔT
Given that the specific heat capacity of water is approximately 4186 J/kg°C:
c = 4186 J/kg°C
The mass of water is not given, but we can assume it is enough to bring the temperature of the iron down. Let's suppose it is 1 kg.
Therefore:
m = 1 kg
ΔT = -50°C
Q = mcΔT
Q = (1 kg)(4186 J/kg°C)(-50°C)
Q = -209,300 J
The negative sign indicates that the water gains heat.
Therefore, the water gains 209,300 Joules of heat.
In summary:
- The hot iron loses 5625 Joules of heat.
- The water gains 209,300 Joules of heat.
heat lost = heat gained.
q = mass x specific heat x delta T.