When a 310-g piece of iron at 180∘C is placed in a 95-g aluminum calorimeter cup containing 250 g of liquid at 10∘C, the final temperature is observed to be 32∘C. The value of specific heat for iron is 450 J/kg⋅C∘, and for aluminum is 900 J/kg⋅C∘. Determine the specific heat of the liquid.

Question

When a 310-g piece of iron at 180∘C is placed in a 95-g aluminum calorimeter cup containing 250 g of liquid at 10∘C, the final temperature is observed to be 32∘C. The value of specific heat for iron is 450 J/kg⋅C∘, and for aluminum is 900 J/kg⋅C∘. Determine the specific heat of the liquid.

Answer 1

To determine the specific heat of the liquid, we can use the principle of energy conservation.

The heat gained by the liquid can be calculated using the equation:

Q1 = mcΔT1

Where:
Q1 is the heat gained by the liquid
m is the mass of the liquid
c is the specific heat of the liquid
ΔT1 is the change in temperature of the liquid

Similarly, the heat lost by the iron can be calculated as:

Q2 = mcΔT2

Where:
Q2 is the heat lost by the iron
m is the mass of the iron
c is the specific heat of iron
ΔT2 is the change in temperature of the iron

According to the principle of energy conservation, the heat gained by the liquid should be equal to the heat lost by the iron. Therefore:

Q1 = Q2

Substituting the values, we have:

mcΔT1 = mcΔT2

Canceling the common terms, we get:

ΔT1 = ΔT2

Therefore, the change in temperature of the liquid is equal to the change in temperature of the iron.

Now, let's calculate the change in temperature of the liquid:

ΔT1 = Final temperature − Initial temperature

ΔT1 = 32∘C − 10∘C = 22∘C

Since the change in temperature of the liquid is the same as the change in temperature of the iron, we can use the same value for ΔT2.

Now, substituting the given values:

22∘C = (mliquid)(specific heat of liquid)

To find the specific heat of the liquid, we need to calculate the mass of the liquid, which can be obtained using the conservation of mass:

Mass of liquid + Mass of iron = Mass of calorimeter cup

Given:
Mass of iron (miron) = 310 g
Mass of calorimeter cup (mcalorimeter) = 95 g
Mass of liquid (mliquid) = Mass of calorimeter cup − Mass of iron

Substituting the given values:

mliquid = 95 g − 310 g = -215 g

Since the mass of the liquid cannot be negative, there seems to be an error in the question. Please double-check the given values.