Q4. A 0.150 kg copper block at 100 °C (s = 386.4 J kg K) is dropped into 0.100 kg of a liquid at 30.0 °C in a calorimeter whose heat capacity is 40 J K. The final temperature is 50.0 °C. Find the specific heat of the liquid.
Heat lost by the block = 0.150 × 386.4 × 50, about 2898 J. The calorimeter gains 40 × 20 = 800 J, so the liquid gains about 2098 J. Then sl = 2098/(0.100 × 20), about 1049 J kg K.
Q5. Assume each thermometer reading has an error bound of ±0.5 °C, and the measured water temperature rise is 5.0 °C. Roughly how large can the error in s be from this reading alone, and what would you change?
Each of the two readings may be out by 0.5 °C, so the rise may be out by up to 1.0 °C, that is 20%. Use a calibrated thermometer with a smaller reading uncertainty, and a larger solid or less water, so that the rise is bigger.
Q6. Why is the outside of the calorimeter polished, and why is it kept inside a jacket with an air gap or glass wool?
A polished surface is a poor radiator, so the calorimeter loses little heat by radiation. A narrow air gap and glass wool reduce conduction and suppress bulk air circulation, so little heat leaks out by conduction and convection. Together these keep the system close to isolated, which the principle of mixtures needs.