Question

A 245.7g sample of metal at 75.2 degrees Celsius was placed in 115.43g water at 22.6

degrees Celsius. The final temperature of the water and metal was 34.6 Celsius. If no heat

was lost to the surroundings what is the specific heat of the metal?

Answer 1

We can use the equation:

q = m1c1ΔT1 = -m2c2ΔT2

where q is the heat transferred, m1 and c1 are the mass and specific heat of the metal, ΔT1 is the change in temperature of the metal, m2 and c2 are the mass and specific heat of the water, and ΔT2 is the change in temperature of the water.

We can solve for the specific heat of the metal, c1, by rearranging the equation:

c1 = -m2c2ΔT2 / (m1ΔT1)

Plugging in the given values, we get:

c1 = -(0.11543 kg)(4.184 J/g°C)(34.6°C - 22.6°C) / (0.2457 kg)(75.2°C - 34.6°C)

c1 = 0.418 J/g°C

Therefore, the specific heat of the metal is 0.418 J/g°C.