Question 26 (1 point) 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 6.47times 10^5 J. exothermic 8.17times 10^4 , exothermic 6.47times 10^5 J, endothermic 8.17times 10^4 , endothermic

The heat transferred when 250 g of water at 98 degrees Celsius cools to room temperature at 20.0 degrees Celsius is $\boxed{8.17\times 10^{4} \text{ J}}$ in an $\boxed{\text{exothermic}}$ process.<br /><br />**Explanation:**<br /><br />1. **Calculate the change in temperature:**<br /> ΔT = T_final - T_initial = 20.0°C - 98°C = -78°C<br /><br />2. **Convert mass to kilograms:**<br /> m = 250 g = 0.250 kg<br /><br />3. **Use the specific heat capacity of water:**<br /> The specific heat capacity of water (c) is 4.184 J/g°C or 4184 J/kg°C.<br /><br />4. **Calculate the heat transfer (q) using the formula:**<br /> q = mcΔT <br /> q = (0.250 kg)(4184 J/kg°C)(-78°C)<br /> q = -81732 J <br /> q ≈ -8.17 x 10⁴ J<br /><br />5. **Interpret the sign:**<br /> The negative sign indicates that heat is being *released* by the water, meaning the process is exothermic. Heat flows out of the system (the water) and into the surroundings.<br />