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
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
