# Would the change in enthalpy (ΔH) for the dissolution of urea in water be positive or negative?

To test the properties of a fertilizer, $$15.0\ \mathrm g$$ of urea, $$\ce{NH2CONH2(s)}$$, is dissolved in $$150\ \mathrm{mL}$$ of water in a simple calorimeter. A temperature change from $$20.6\ \mathrm{^\circ C}$$ to $$17.8\ \mathrm{^\circ C}$$ is measured. Calculate the molar enthalpy of solution for the fertilizer urea

I worked through this question by finding $$Q = mc\Delta T$$, and then dividing $$Q$$ by the moles of urea present. I can tell the process is endothermic because $$\Delta T$$ is negative, however my answer for $$\Delta H$$ comes out as negative, which would only make sense if this was an exothermic reaction. I'm not sure where I am wrong to be honest.

Here is my work:

$$\Delta H = (150\ \mathrm{mL} \times 1\ \mathrm{g/mL} \times 4.18\ \mathrm{J/(g\ ^\circ C}) \times -2.8\ \mathrm{^\circ C}) / (15\ \mathrm g / 60.07\ \mathrm g) = -7030.59\ \mathrm{J/mol} = -7.03\ \mathrm{kJ/mol}$$

TL;DR - question asks for $$\Delta H$$ of an endothermic process, not sure if my answer should be positive or negative

The sign of Q depends on the perspective. The water temperature decreased because it "lost" heat. The process of dissolving urea required energy, it "gained" energy. If I give you a penny, should that be +1 or -1 penny? Well, it depends who you ask.

In your answer, you are missing a negative sign in $$\Delta H=−Q$$ the way you start out with $$Q$$ from the perspective of the water.