Answer:
Endothermic reactions are those that absorb heat from their surroundings, while exothermic reactions release heat into their surroundings. In terms of the sign of ΔH, endothermic reactions have a positive ΔH value, while exothermic reactions have a negative ΔH value.
In the given problem, there is no reaction between the metal and water, so we can assume that the temperature change is due to heat transfer from the metal to the water. Since the final temperature is less than the initial temperature of the metal, we know that the process is exothermic, and ΔH is negative.
To calculate the specific heat of the metal, we can use the formula:
q = mcΔT
where q is the heat transferred, m is the mass of the water, c is the specific heat of water, and ΔT is the change in temperature of the water. We can assume that the heat transferred from the metal to the water is equal to the heat lost by the metal, so we can write:
q = -mcΔT
where the negative sign indicates that the heat lost by the metal is equal in magnitude but opposite in sign to the heat gained by the water.
The mass of the water is given as 20.00 g, the specific heat of water is 4.184 J/g°C, and the change in temperature of the water is (26.8 - 22.3) = 4.5°C. Substituting these values into the formula above, we get:
q = -(20.00 g)(4.184 J/g°C)(4.5°C) = -376.8 J
The negative sign indicates that the heat lost by the metal is equal in magnitude but opposite in sign to the heat gained by the water.
The heat lost by the metal is equal to the heat transferred, so:
q = -mcΔT
where m is the mass of the metal, c is the specific heat of the metal, and ΔT is the change in temperature of the metal.
The mass of the metal is given as 25.40 g, and the initial and final temperatures of the metal are 90.0°C and 26.8°C, respectively. Substituting these values into the formula above and solving for c, we get:
c = -q/(mΔT) = -(-376.8 J)/(25.40 g)(90.0°C - 26.8°C) = 0.232 J/g