Specific Heat Capacity: Systematic Errors in Method of Mixtures
A student determines the specific heat capacity of a metal block using the method of mixtures. A 0.200 kg block of the metal is heated in boiling water at 100 °C for several minutes. It is then quickly transferred to a calorimeter containing 0.150 kg of water at 20.0 °C. The calorimeter has a heat capacity of 50 J/K. The mixture reaches a maximum temperature of 28.5 °C. The specific heat capacity of water is 4200 J/(kg K). The student calculates the specific heat capacity and obtains 404 J/(kg K). They repeat the experiment three times and obtain 380, 410, and 395 J/(kg K). The accepted values for some metals are: copper 385, zinc 388, brass 380, iron 450 J/(kg K). During the experiment, the student notices that a few drops of water splashed out during transfer, and the thermometer reading lagged behind the actual temperature. They also did not account for heat loss to the surroundings.
1. Verify the student's calculation of 404 J/(kg K) using the given data. Show your working.
Read the worked answer
Using conservation of energy: heat lost by the metal = heat gained by the water + heat gained by the calorimeter.
Let c_m be the specific heat capacity of the metal. The metal cools from 100 °C to 28.5 °C, so ΔT_m = 71.5 K. The water warms from 20.0 °C to 28.5 °C, so ΔT_w = 8.5 K.
Heat gained by water: Q_w = m_w c_w ΔT_w = 0.150 × 4200 × 8.5 = 5355 J. Heat gained by calorimeter: Q_cal = C_cal ΔT_w = 50 × 8.5 = 425 J. Total heat gained = 5355 + 425 = 5780 J.
Heat lost by metal: Q_m = m_m c_m ΔT_m = 0.200 × c_m × 71.5 = 14.3 c_m.
Equating: 14.3 c_m = 5780, so c_m = 5780 / 14.3 = 404.2 J/(kg K). The student's value of 404 J/(kg K) is correct.
2. Discuss how the splashing of water and thermometer lag would affect the calculated specific heat capacity (increase, decrease, or uncertain), and explain your reasoning.
Read the worked answer
Splashing of water: If water splashes out during transfer, the actual mass of water in the calorimeter is less than the 0.150 kg used in the calculation. The calculation therefore overestimates the heat gained by the water (since it uses too large a mass). This leads to an overestimate of the heat lost by the metal, and hence an overestimate of c_m. Thus, splashing causes the calculated specific heat capacity to be too high.
Thermometer lag: If the thermometer reading lags behind the actual temperature, the recorded maximum temperature is lower than the true final temperature. This means the recorded ΔT_w is smaller than the true ΔT_w, and the recorded ΔT_m is larger than the true ΔT_m. In the formula c_m = (m_w c_w + C_cal) ΔT_w / (m_m ΔT_m), a smaller numerator and a larger denominator both lead to a smaller calculated c_m. Thus, thermometer lag causes an underestimate of c_m.
Heat loss to surroundings: Some heat from the metal goes to the surroundings instead of the water, so the water temperature rise is smaller than it would be otherwise. This also leads to an underestimate of c_m. The net effect of these systematic errors is uncertain because splashing increases c_m while thermometer lag and heat loss decrease it.
3. Given the repeated values, determine the most likely metal and justify your choice, considering the systematic errors.
Read the worked answer
The four values obtained are 404, 380, 410, and 395 J/(kg K). The mean is (404 + 380 + 410 + 395) / 4 = 397.25 J/(kg K). The accepted values are: copper 385, zinc 388, brass 380, iron 450 J/(kg K). The mean is closest to zinc (388) and copper (385), with a difference of about 9–12 J/(kg K). Brass (380) is 17 J/(kg K) lower, and iron (450) is far too high.
Considering the systematic errors: splashing causes an overestimate, while thermometer lag and heat loss cause underestimates. If splashing is significant, the true value could be lower than 397, bringing it closer to copper (385) or zinc (388). If thermometer lag and heat loss dominate, the true value could be even lower, which would favour copper or brass. However, the scatter of the repeated values (±15 J/(kg K)) suggests experimental uncertainty. The most likely metal is copper (385 J/(kg K)) because the mean is slightly above the accepted value, and the overestimate from splashing could explain the difference. Zinc (388 J/(kg K)) is also a strong candidate. Brass (380) is less likely because the mean is noticeably higher. Iron (450) is ruled out. The student should improve the experiment by minimizing splashing, using a more responsive thermometer, and insulating the calorimeter to reduce heat loss.