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 = qmetal

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 = qmetal

anonymous · Answer

attachment

anonymous · Answer

U mean the energy change?

anonymous · Answer

I believe that's what they're asking. 

I so far have: 
$$q=27.776 * 4.18 * ?$$

I just don't know how to find the last part of the equation and I don't know if I'm doing this correct or not.

anonymous · Answer

Energy gained by water = Energy lost by metal
Energy gained by water =  $$m \times c \times \theta$$
= 27.776 * 4.18 * (38.9-25.3)

The temperature at equilibrium ( mixture is 38.9)

anonymous · Answer

So q is 1579?

anonymous · Answer

yes if the calculation is right :)

anonymous · Answer

and that is all? wow ok thank you!

anonymous · Answer

:)