The heat transferred when 250 g of water at 98 degrees Celsius cools to room temperature at 20.0 degrees Celsius is __ in an __ process

The heat transferred when 250 g of water at 98 degrees Celsius cools to room temperature at 20.0 degrees Celsius is **81.2 kJ** in an **exothermic** process.<br /><br />Here's the solution:<br /><br />1. **Formula:** We use the formula for heat transfer: q = mcΔT<br /><br /> Where:<br /> * q = heat transferred (in Joules)<br /> * m = mass of the substance (in grams)<br /> * c = specific heat capacity of the substance (for water, it's 4.184 J/g°C)<br /> * ΔT = change in temperature (in °C)<br /><br />2. **Calculate ΔT:** ΔT = Final Temperature - Initial Temperature = 20.0°C - 98°C = -78°C<br /><br />3. **Plug in the values:** q = (250 g) * (4.184 J/g°C) * (-78°C) = -81192 J<br /><br />4. **Convert to kJ:** -81192 J * (1 kJ / 1000 J) = -81.192 kJ (approximately -81.2 kJ)<br /><br />5. **Exothermic Process:** The negative sign indicates that heat is being *released* by the water, meaning it's an exothermic process. Cooling is always exothermic.<br />