Question

84.g of a metal are heated to 100°c, an then placed in a coffee cup calorimeter containig 50.0g of water at 20.0°c the final temp in the caloriomete is 50°c what is the specific heat capacity?

Answer 1

To calculate the specific heat capacity of the metal, use the equation q(metal) + q(water) = 0. Plug in the given values and solve for c(metal).

Step-by-step explanation:

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

q(metal) + q(water) = 0

We know that the metal gains heat and the water loses heat, so we can write the equation as:

m(metal) * c(metal) * ( Ti) + m(water) * c(water) * ( Ti) = 0

Plugging in the values:

40.0g * c(metal) * (60.15°C - 130.0°C) + 50.0g * 4.18 J/g°C * (60.15°C - 20.0°C) = 0

Simplifying the equation, we can solve for c(metal):

40.0g * c(metal) * (-69.85°C) + 50.0g * 4.18 J/g°C * 40.15°C = 0

c(metal) = (50.0g * 4.18 J/g°C * 40.15°C) / (40.0g * -69.85°C)

c(metal) = -0.342 J/°C*g