Question

30. g of an unknown substance at 120.0∘C is placed in 600.0 g of water at 30.0∘C. The final temperature of the water is 31.4∘C. What was the specific heat of the metal?

Round your answer to one decimal place

Use 4.184Jg∘C for the specific heat of water.

Answer 1

The specific heat capacity of the unknown metal is calculated using the heat transfer between the metal and water and applying the principle of conservation of energy. After the calculations, the specific heat capacity of the unknown metal is approximately 1.32 J/g°C.

Step-by-step explanation:

To calculate the specific heat capacity of the unknown metal, we can use the principle of conservation of energy, which implies that the heat lost by the metal will be equal to the heat gained by the water. Using the specific heat equation, q = mcΔT (where q is the heat energy, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature), we can set up two equations, one for the water and one for the metal:

For water: q_water = m_water * c_water * ΔT_water

For metal: q_metal = m_metal * c_metal * ΔT_metal

Since the heat lost by the metal is gained by the water, we have q_metal = -q_water.

Plugging in the known values:

• m_water = 600.0 g
• c_water = 4.184 J/g°C
• ΔT_water = 31.4°C - 30.0°C = 1.4°C

Calculate q_water:

q_water = 600.0 g * 4.184 J/g°C * 1.4°C = 3514.56 J

For the metal:

• m_metal = 30.0 g
• ΔT_metal = 120.0°C - 31.4°C = 88.6°C

Since q_metal = -q_water, we can write:

-3514.56 J = 30.0 g * c_metal * -88.6°C

Solving for c_metal, we get:

c_metal = 3514.56 J / (30.0 g * 88.6°C)

c_metal therefore equals approximately 1.32 J/g°C when rounded to one decimal place.