Question

A tin calorimeter cup with a specific heat of 210 J/(kg · °C) contains milk with a specific heat of 4,186 J/(kg · °C) at 32.0°C. The combination is heated uniformly so that the temperature changes by 1.30°C per minute. If the cup has a mass of 100 g and contains 900 g of the liquid, determine the rate at which heat energy is added (in W).

Answer 1

To determine the rate at which heat energy is added to the system, we need to calculate the rate of change of thermal energy within the system.

This can be done using the formula:

Rate of Heat Energy Added=Mass×Specific Heat×Temperature Change Rate

Rate of Heat Energy Added=Mass×Specific Heat×Temperature Change Rate.

First, we'll calculate the total mass of the system, which is the sum of the mass of the cup and the mass of the milk:

Total mass = Mass of cup + Mass of milk

Total mass = 100 g + 900 g

Total mass = 1000 g = 1 kg.

Now we'll calculate the rate of heat energy added:

Rate of Heat Energy Added = Total mass × Specific heat of the mixture × Temperature change rate

Rate of Heat Energy Added = 1 kg × (210 J/(kg·°C) + 900 g × 4,186 J/(kg·°C)) × 1.30°C/min

Rate of Heat Energy Added = 1 kg × (210 J/(kg·°C) + 376914 J/(kg·°C)) × 1.30°C/min

Rate of Heat Energy Added = 1 kg × 377124 J/(kg·°C) × 1.30°C/min

Rate of Heat Energy Added = 491061.2 J/min.

Since 1 W (Watt) is equivalent to 1 J/s (Joule per second), we can convert the rate of heat energy from J/min to W by dividing by 60 (since there are 60 seconds in a minute):

Rate of Heat Energy Added = 491061.2 J/min ÷ 60 s/min

Rate of Heat Energy Added ≈ 8184.35 W.

Therefore, the rate at which heat energy is added to the system is approximately 8184.35 watts (W).