Question

A copper forceps with a mass of 500 g was accidentally left on a heating mantle until it reached a temperature of 35 °C. Subsequently, a student placed the forceps in 70 g of water at 22 °C to cool it off. Calculate the specific heat capacity of the copper forceps if the final temperature of the water was 27 °C. (Specific heat capacity of water is 4.18 J/°C.g)

Answer 1

The specific heat capacity of the copper forceps is 4.18 J/°C.g.

Step-by-step explanation:

To calculate the specific heat capacity of the copper forceps, we can use the equation:

q = mcΔT

where q is the heat transferred, m is the mass of the object, c is the specific heat capacity, and ΔT is the change in temperature.

Substituting the given values, we have:

q = mcΔT

q = (500 g + 70 g)(4.18 J/°C.g)(27 °C - 22 °C)

q = 570 g * 4.18 J/°C.g * 5 °C

q = 11919 J

Since all the heat transferred from the copper forceps is absorbed by the water, we can equate q to the heat gained by the water:

q = mcΔT

11919 J = (70 g)(4.18 J/°C.g)(27 °C - 22 °C)

11919 J = 70 g * 4.18 J/°C.g * 5 °C

Solving for c, we get:

c = q / (mΔT)

c = 11919 J / (570 g * 5 °C)

c = 4.18 J/°C.g

Therefore, the specific heat capacity of the copper forceps is 4.18 J/°C.g.