The specific heat capacity of the zinc, given that 18.3 grams of zinc at 98.19 °C is dropped into a cup containing water at 20.75 °C is 0.374 J/gºC
How to calculate the specific heat capacity of the zinc?
We shall use the conservation of heat energy to solve the quuestion given above.
First, we shall determine the heat the water absorbed when the zinc was dropped into it. This is shown below:
- Mass of water (M) = 79.6 grams
- Initial temperature (T₁) = 20.75 °C
- Final temperature (T₂) = 22.31 °C
- Change in temperature (ΔT) = 22.31 - 20.75 = 1.56 °C
- Specific heat capacity of water (C) = 4.184 J/gºC
- Heat absorbed (Q) =?
Q = MCΔT
= 79.6 × 4.184 × 1.56
= 519.552384 J
Haven obtain the heat the water absorbed when the zinc was dropped into it, we shall calculate the specific heat capacity of the zinc. Details below:
- Heat absorbed by water when the zinc was dropped (Q) = 519.552384 J
- Heat released by zinc in the process (Q) = -519.552384 J
- Mass of zinc (M) = 18.3 grams
- Initial temperature of zinc (T₁) = 98.19 °C
- Final temperature of zinc (T₂) = 22.31 °C
- Change in temperature of zinc (ΔT) = 22.31 - 98.19 = -75.88 °C
- Specific heat capacity of zinc (C) = ?
Q = MCΔT
-519.552384 = 18.3 × C × -75.88
-519.552384 = -1388.604 × C
C = -519.552384 / -1388.604
= 0.374 J/gºC
Complete question:
She heats 18.3 grams of zinc to 98.19 °C and then drops it into a cup containing 79.6 grams of water at 20.75 °C. She measures the final temperature to be 22.31 °C.
Assuming that all of the heat is transferred to the water, she calculates the specific heat of zinc to be ___J/g °C.