A 360.0 g steel rebar (0.449 J/g degree C) is dropped into 425 mL of water at 24.0 degrees C, giving a final 42.7 degrees C. What was the rebar starting temperature?

1. Water density is 1.0 g/mL, so 425 mL of water has a mass of 425 g. 2. Heat taken in by the water is 4.184 J/g degree C times 425 g times (42.7 minus 24.0). 3. That is 1778.2 times 18.7, which equals 33,252 J. 4. The rebar gave off that heat, so its heat value is minus 33,252 J. 5. For the rebar, 0.449 J/g degree C times 360.0 g gives 161.64 J per degree Celsius. 6. Its temperature change is minus 33,252 divided by 161.64, which is minus 205.7 degrees Celsius. 7. The final temperature is 42.7 degrees C, so the start was 42.7 plus 205.7, about 248 degrees C. 8. Answering 206 degrees C reports the temperature drop rather than the starting temperature. _Source: OpenStax Chemistry (CC BY 4.0), Ch 5 "Thermochemistry", section 5.2 Calorimetry_