Question

So I was able to figure out the first part of the question ---

"Calculate the energy change (q) of the surroundings (water) using the enthalpy equation

qwater = m × c × ΔT.

We can assume that the specific heat capacity of water is 4.18 J / (g × °C) and the density of water is 1.00 g/mL.

The water has absorbed the heat of the metal. So, qwater = qunknown metal
Qwater=25*4.18J/(°C*g)*(38.7°C-25°C)
Multiply 25 by 4.18 (mass of the metal by the SHC of water)
Qwater= 104.18/(°C*g)*(7.6°C)
Multiply 104.18 by 7.6 (joules by temperature)
Qwater=794.2 /(°C*g)"

Answer 1

Using the formula qunknown metal = m x c x (triangle)T, calculate the specific heat of the metal. Use the data from your experiment for the unknown metal in your calculation. Show ALL your work.

Answer 2

can you post the initial data (e.g. mass of metal, mass of water), i find what you wrote sort of confusing.

Answer 3

If you found the change in energy of the water, we know that the change in energy of the metal MUST BE the exact same. So you have:\[Q_{water} = m_{water}*C_{water}*\Delta T_{water} = 794 Joules\]

Answer 4

use that amount of energy in a NEW enthalpy equation, but instead of solving for Q, you're solving for C, because now you're solving for the METAL, not water\[Q_{metal} = m_{metal}*C_{metal}*\Delta T_{metal}\]