Question

A copper pot with a mass of 2 kg is sitting at room temperature (20 C) a. If 200 g of boiling water (100 C) are put in the pot, after a few minutes the water and pot come to the same temperature. What temperature is this? Specific Heat Thermal Conductivity (J/kg/C) 385 4182 (W/m2/C) 375 0.6 Copper Water/

Answer 1

To find the specific heat capacity of a metal, the heat lost by the metal equals the heat gained by the water it's placed in. We use the formula Q=mcΔT and set the heat lost and gained equal to each other to solve for the metal's specific heat capacity.

Step-by-step explanation:

To determine the specific heat capacity of the metal in the given scenario, where 34.5 grams of hot metal at 75°C is placed in 64.0 grams of water at 25°C and they reach thermal equilibrium at 39°C, we use the principle of conservation of energy. The heat lost by the metal will equal the heat gained by the water.

The formula for the heat change is:

Q = mcΔT

where

• Q is the heat absorbed or released,
• m is the mass of the substance,
• c is the specific heat capacity,
• ΔT is the change in temperature.

For the water, we have:

Qwater = (64.0 g)(4.18 J/g°C)(39°C - 25°C)

For the metal, assuming its specific heat is 'c':

Qmetal = (34.5 g)(c)(39°C - 75°C)

Since Qmetal equals -Qwater (heat lost by metal is equal to heat gained by water), we can set them equal and solve for 'c':

(34.5 g)(c)(-36°C) = -(64.0 g)(4.18 J/g°C)(14°C)

Now, solve for 'c' to find the specific heat capacity of the metal.