Question

A 58.5 g sample of glass is put into a calorimeter (see sketch at right) that contains 250.0 g of water. The glass sample starts off at 91.2 °C and the temperature of the water starts off at 19.0 °C when the temperature of the water stops changing it's 21.7 The pressure remains constant at 1 atm. Calculate the specific heat capacity of glass according to this experiment. Be sure your answer is rounded to 2 significant digits

Answer 1

Answer:

The specific heat of the glass = 0.70 J/g°C

Explanation:

Step 1: Data given

Mass of glass = 58.5 grams

Mass of water = 250.0 grams

Initial temperature of glass = 91.2 °C

Initial temperature of water = 19.0 °C

Final temeprature = 21.7°C  

Pressure = 1 atm

Specific heat capacity of water = 4.184 J/g°C

Step 2: Calculate the specific heat of glass

Heat gained = heat lost

Qwater = -Qglass

Q = m*c*ΔT

m(water) * c(water) * ΔT = -m(glass) * c(glass) * ΔT(glass)

⇒ with mass of water = 250.0 grams

⇒ with specific heat of water = 4.184 J/g°C

⇒ with ΔT = T2 - T1 = The change in temperature = 21.7 - 19.0 = 2.7 °C

⇒ with mass of glass = 58.5 grams

⇒ with specific heat of glass = ?

⇒ with ΔT = T2 - T1 = The change in temperature = 21.7 - 91.2 °C = -69,5 °C

250*4.184*2.7 =-58.5*C(glass) * -69.5

2824.2 =4065.75*C(glass)

C(glass) = 0.70 J/g°C

The specific heat of glass = 0.70 J/g°C

Answer 2

Answer:

The specific heat capacity of glass is 0.70J/g°C

Explanation:

Heat lost by glass = heat gained by water

Heat lost by glass = mass × specific heat capacity (c) × (final temperature - initial temperature) = 58.5×c×(91.2 - 21.7) = 4065.75c

Heat gained by water = mass × specific heat capacity × (final temperature - initial temperature) = 250×4.2×(21.7 - 19) = 2835

4065.75c = 2835

c = 2835/4065.75 = 0.70J/g°C