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Error Analysis 1. In parts A, B, and C, if the calorimeter was not dry, the change in temperature would be too (high/low). This would make the |AH| appear too _(high/low). 2. If a reaction is exothermic, the sign of AH is and heat is 3. If a reaction is endothermic, the sign of AH is and heat is 4. In the determination of the specific heat of the metal, if some of the water that was used to heat the metal was not removed before the metal was transferred to the calorimeter, the final temperature would be too (high/low). This would make the specific (high/low). Hint: put in hypothetical numbers.

Answer 1

1) if the calorimeter is not dry... Water will also absorb some heat and thus the change in the temperature will be low and thus change in enthalpy ( delta H) will also be low.

2) if reqction is endothermic sign of $\Delta$ H is positive and heat is absorbed

3) if reaction is exothermic sign of $\Delta$ H is negative and heat is released.

4) while heating the metal, water droplets around the metal will also get heated and thus change in temperature of the metal will be low and thus the specufic heat will recorded to be high.