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Answer :
The specific heat capacity of the metal, given that 27.777 g of water at 22.1 °C was mixed with the metal is 9.7×10⁻² Cal/gºC
How do I determine the specific heat capacity of the metal?
Step 1: Obtain the heat absorbed by the water. This is shown below:
- Mass of water (M) = 27.777 g
- Initial temperature (T₁) = 22.1 °C
- Final temperature (T₂) = 29.3 °C
- Temperature change (ΔT) = 29.3 - 22.1 = 7.2 °C
- Specific heat capacity of water (C) = 1 Cal/gºC
- Heat absorbed (Q) =?
Q = MCΔT
= 27.777 × 1 × 7.2
= 199.9944 Cal
Step 2: Determine the specific heat capacity of the metal using the heat absorbed by the water. Details below:
- Heat absorbed by water (Q) = 199.9944 Cal
- Heat released by metal (Q) = -199.9944 Cal
- Mass of metal (M) = 29.292 g
- Initial temperature (T₁) = 99.9 °C
- Final temperature (T₂) = 29.3 °C
- Temperature change (ΔT) = 29.3 - 99.9 = -70.6 °C
- Specific heat capacity of metal (C) = ?
Q = MCΔT
-199.9944 = 29.292 × C × -70.6
-199.9944 = -2068.0152 × C
Divide both sides by -2068.0152
C = -199.9944 / -2068.0152
= 9.7×10⁻² Cal/gºC
Thus, the specific heat capacity of the metal is 9.7×10⁻² Cal/gºC. None of the options are correct.
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