A student heats a 100.0 g sample of liquid water in a beaker. The temperature of the water increases from 22.0°C to 28.0°C. Assume the specific heat capacity of water is . What is the heat transferred to the water, ?
- +2,510 J (correct answer)
- +251 J
- -2,510 J
- +418 J
- +10,000 J
Explanation: This question tests the skill of heat capacity and calorimetry. The heat transferred to the water is calculated using the formula q = m c ΔT, where m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. Here, the system is the water sample, which absorbs heat to increase its temperature from 22.0°C to 28.0°C, resulting in a positive ΔT of 6.0°C. Plugging in the values, q = 100.0 g × 4.18 J g⁻¹ °C⁻¹ × 6.0°C = +2,510 J, correctly modeling the energy absorbed by the water. A tempting distractor is choice C, -2,510 J, which arises from the misconception of assigning a negative sign to q when the system gains heat, confusing the sign convention. Always calculate ΔT as T_final - T_initial and assign the sign of q based on whether the system absorbs (positive) or releases (negative) heat.