Question

A cool water sample absorbed 1,710 J of energy from some metal. The temperature of the 59.0 g piece of metal changed from 99.5 °C to 27.8 °C. What is the specific heat of the metal? 9₂H₂0 = 1, 710 J c,metal (J/g °C) Cₘₑₜₐₗ = [?] J/g.ºC

remember qₘₑₜₐₗ = -q,H₂O

Answer 1

The specific heat of the metal is calculated using the formula c_metal = -q_H2O / (m_metalΔT_metal). By substituting the values, the specific heat is found to be 0.404 J/g°C.

Step-by-step explanation:

To find the specific heat of the metal that changed from 99.5 °C to 27.8 °C and absorbed 1,710 J of energy, we can use the formula q = mcΔT, where q is the heat transferred, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. Given that the heat lost by the metal is equal to the heat gained by the water (qmetal = -qH2O), we can rearrange the formula to solve for the metal's specific heat (cmetal):

cmetal = -qH2O / (mmetalΔTmetal)

Plugging in the given values:

cmetal = -1,710 J / (59.0 g × (27.8 °C - 99.5 °C))

cmetal = -1,710 J / (59.0 g × (-71.7 °C))

cmetal = -1,710 J / (-4,229.3 g°C) = 0.404 J/g°C

Therefore, the specific heat of the metal is 0.404 J/g°C.