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Difficulty: Very hardHeat Capacity and Specific Heat Capacity

A copper calorimeter of heat capacity 300 J K−1300\text{ J K}^{-1} contains 0.5 kg0.5\text{ kg} of water at an initial temperature of 25∘C25^\circ\text{C}. An electric heater rated at 800 W800\text{ W} is immersed in the water to heat the system for 4 minutes4\text{ minutes}. If heat is lost to the surrounding environment at a constant rate of 200 W200\text{ W} throughout the heating duration, what is the final temperature of the water-calorimeter system in ∘C^\circ\text{C}? (Take the specific heat capacity of water as 4200 J kg−1K−14200\text{ J kg}^{-1}\text{K}^{-1})

Answer: 85 °C

Answer

The final temperature of the system is 85∘C85^\circ\text{C}.
The net thermal energy added to the system accounts for both the supplied electrical energy and the heat energy lost to the surroundings: Qnet=(800−200) W×240 s=144,000 JQ_{\text{net}} = (800 - 200)\text{ W} \times 240\text{ s} = 144,000\text{ J}. The total heat capacity of the water-calorimeter system is Ctotal=300 J K−1+(0.5 kg×4200 J kg−1K−1)=2400 J K−1C_{\text{total}} = 300\text{ J K}^{-1} + (0.5\text{ kg} \times 4200\text{ J kg}^{-1}\text{K}^{-1}) = 2400\text{ J K}^{-1}. The temperature increase is ΔT=144,0002400=60∘C\Delta T = \frac{144,000}{2400} = 60^\circ\text{C}. Adding this to the initial temperature of 25∘C25^\circ\text{C} gives the final temperature of 85∘C85^\circ\text{C}.

Step-by-Step Solution

1
Convert heating time to standard SI units (seconds)
t=4×60 s=240 st = 4 \times 60\text{ s} = 240\text{ s}
Power is measured in Joules per second (Watts), so time must be in seconds.
2
Calculate the net rate of heat energy input to the system
Pnet=800 W−200 W=600 WP_{\text{net}} = 800\text{ W} - 200\text{ W} = 600\text{ W}
The net heating power is the input power minus the power dissipated as heat loss.
3
Calculate total net heat energy transferred to the system
Qnet=Pnet×t=600 W×240 s=144,000 JQ_{\text{net}} = P_{\text{net}} \times t = 600\text{ W} \times 240\text{ s} = 144,000\text{ J}
Thermal energy transferred equals net power multiplied by time.
4
Compute the total heat capacity of the combined system (water + calorimeter)
Ctotal=Ccalorimeter+(mwater×cwater)=300 J K−1+(0.5 kg×4200 J kg−1K−1)=2400 J K−1C_{\text{total}} = C_{\text{calorimeter}} + (m_{\text{water}} \times c_{\text{water}}) = 300\text{ J K}^{-1} + (0.5\text{ kg} \times 4200\text{ J kg}^{-1}\text{K}^{-1}) = 2400\text{ J K}^{-1}
Heat capacity of water is mass multiplied by specific heat capacity, added to the calorimeter's given heat capacity.
5
Calculate the temperature rise of the system
ΔT=QnetCtotal=144,000 J2400 J K−1=60∘C\Delta T = \frac{Q_{\text{net}}}{C_{\text{total}}} = \frac{144,000\text{ J}}{2400\text{ J K}^{-1}} = 60^\circ\text{C}
Temperature change is total net heat supplied divided by total heat capacity.
6
Find the final temperature of the system
Tfinal=Tinitial+ΔT=25∘C+60∘C=85∘CT_{\text{final}} = T_{\text{initial}} + \Delta T = 25^\circ\text{C} + 60^\circ\text{C} = 85^\circ\text{C}
Final temperature equals initial temperature plus temperature increase.

Key Concept

Conservation of thermal energy in calorimeter systems with continuous power loss
Estimated Time:1m 30s
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Topics

Heat Capacity and Specific Heat Capacity

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