The 836 is ok; you just didn't go far enough.
q = heat water absorbed + heat cal absorbed.
1000 = [mass H2O x sp.h. H2O x (Tfinal-Tinitial) + Qcal
1000 = (50 x 4.18 x 4) + Qcal
1000 = 836 + Qcal
Qcal = 1000-836 = 164
The answer is 164J. How do you get this? I tried Mass x specific heat x delta T and got 836 J but this is wrong.
q = heat water absorbed + heat cal absorbed.
1000 = [mass H2O x sp.h. H2O x (Tfinal-Tinitial) + Qcal
1000 = (50 x 4.18 x 4) + Qcal
1000 = 836 + Qcal
Qcal = 1000-836 = 164
First, let's calculate the heat lost by the hot metal using the formula you mentioned:
Heat lost by the metal = mass × specific heat × delta T
Since the specific heat of water is given, we can use it to find the mass of the water in the calorimeter. The mass of water can be found by multiplying the volume of water (50 mL) by its density, which is approximately 1 g/mL.
Mass of water = volume of water × density of water
Mass of water = 50 mL × 1 g/mL
Mass of water = 50 g
We know that the initial temperature of the water is 22°C and the final temperature (equilibrium) is 26°C. Therefore, the change in temperature, delta T, would be:
Delta T = final temperature - initial temperature
Delta T = 26°C - 22°C
Delta T = 4°C
Now, we can calculate the heat lost by the hot metal:
Heat lost by the metal = mass of metal × specific heat of metal × delta T
Heat lost by the metal = 1000 J
Since the heat lost by the metal is equal to the heat gained by the calorimeter:
Heat gained by the calorimeter = heat lost by the metal
Heat gained by the calorimeter = 1000 J
Therefore, the heat gained by the calorimeter is 1000 J.
I apologize for the confusion earlier. The correct answer is actually 1000 J, not 164 J.