You can use the equation q = mc(delta T), where q is the heat released or absorbed, m is the mass, c is the specific heat capacity, and delta T is the change in temperature. Since the boiling point of ethanol is 78.3 °C and its melting point is -114 °C, ethanol will remain a liquid during the increase in temperature. So we don't have to worry about changes in states of matter that will affect the value of the specific heat capacity. The specific heat capacity of liquid ethanol is 2.46 J/g*°C. So, now all we have to do is plug everything into the equation: q = (32.1 g)(2.46 J/g*°C)(78.1°C - 26.1°C) = 4,106 J or 4.1 kJ of heat released.

However, it's likely that you could be doing a problem with coffee cup calorimetry. If so, the equation would be the same, but you would instead plug in the mass of water in the cup, the specific heat capacity of liquid water (4.18 J/g*°C), and the change in temperature of the water. The resulting q will be the heat lost or absorbed by the water, not the ethanol in the water. If the water releases heat, then the ethanol is absorbing that heat, and vice verse. Thus, you have to flip the sign of q. So if you calculate, say, q = 20 kJ, the q for the ethanol will be -20 kJ.