Question

A 12.5-g piece of glass at 65.3 °C is dropped into 45 g of water at 22.5 °C. All the heat lost by the glass is used to heat the water, and the final temperature of the water and glass is 25.8 °C. Calculate the specific heat, in J/g °C, of the glass.

Answer 1

The specific heat of the glass is approximately 5.25 J/g °C.

Step-by-step explanation:

To calculate the specific heat of the glass, we can use the formula for heat transfer: Q = mcΔT Where Q is the heat transferred m is the mass of the glass c is the specific heat of the glass ΔT is the change in temperature of the glass Since all the heat lost by the glass is used to heat the water, the heat transferred by the glass is equal to the heat gained by the water. Therefore, we can set up the equation: mcΔT = mwCwΔT Where mw is the mass of the wate Cw is the specific heat of water ΔT is the change in temperature of the water Given m = 12.5 g ΔT = 65.3 °C - 25.8 °C = 39.5 °C mw = 45 g ΔT = 22.5 °C - 25.8 °C = -3.3 °C Cw = 4.184 J/g °C Plugging in the values, we can solve for c: (12.5 g (c)(39.5 °C) = (45 g)(4.184 J/g °C)(-3.3 °C) c = (-3.3 °C)(45 g)(4.184 J/g °C) / ((12.5 g)(39.5 °C)) = -5.2536 J/g °C ≈ -5.25 J/g °CSince specific heat should be positive, the specific heat of the glass is approximately 5.25 J/g °C.