Question

A 53.0-g metal weight, heated to 84.00°C, is placed into 175 g of water at 20.50°C contained in a perfectly insulating thermos flask. After some time, the temperature inside the thermos flask stabilizes at 23.65°C. The specific heat capacity of water is approximately 4.18 J/K/g in the temperature range 16°C - 61°C. Calculate the specific heat capacity of the metal.

Answer 1

0.44 J/g°C

Step-by-step explanation:

Use the formula:

Q = mcΔT

Where:

Q is the heat energy (J),

m is the mass (g),

c is the specific heat capacity (J/g°C), and

ΔT is the change in temperature (°C).

1. Calculate the heat lost by the metal:

Q_metal = m_metal x c_metal x ΔT_metal

Where:

m_metal = 53.0 g (mass of the metal)

ΔT_metal = 84.00°C - 23.65°C (initial temperature - final temperature)

2. Calculate the heat gained by the water:

Q_water = m_water x c_water x ΔT_water

Where:

m_water = 175 g (mass of the water)

ΔT_water = 23.65°C - 20.50°C (final temperature - initial temperature)

The specific heat capacity of water is approximately 4.18 J/g°C in this temperature range.

Now, since the heat lost by the metal is gained by the water, u can equate these two equations:

Q_metal = Q_water

m_metal x c_metal x ΔT_metal = m_water x c_water x ΔT_water

Now, u can solve for the specific heat capacity of the metal, c_metal:

c_metal = (m_water x c_water x ΔT_water) / (m_metal x ΔT_metal)

Substitute the known values:

c_metal = (175 g x 4.18 J/g°C x (23.65°C - 20.50°C)) / (53.0 g x (84.00°C - 23.65°C))

Calculate the value of c_metal:

c_metal ≈ 0.44 J/g°C

So, the specific heat capacity of the metal is approximately 0.44 J/g°C.