A cube at 112 °C radiates heat at a rate of 92 J/s. If its surface temperature is increased to 196 °C, the rate at which it will now radiate is closest to
The Correct Answer and Explanation is:
The rate at which an object radiates heat is governed by the Stefan-Boltzmann law, which states:P=σAT4P = \sigma A T^4P=σAT4
Where:
- PPP is the power radiated (in watts or joules per second, J/s),
- σ\sigmaσ is the Stefan-Boltzmann constant (5.67×10−8 W/m2K4)(5.67 \times 10^{-8} \, \text{W/m}^2 \text{K}^4)(5.67×10−8W/m2K4),
- AAA is the surface area of the object,
- TTT is the absolute temperature in Kelvin (K).
Key Idea:
- The rate of heat radiation is proportional to the fourth power of the temperature.
- The temperatures need to be converted into Kelvin before using the Stefan-Boltzmann law.
Given:
- Initial temperature: T1=112∘C=112+273.15=385.15 KT_1 = 112^\circ C = 112 + 273.15 = 385.15 \, KT1=112∘C=112+273.15=385.15K
- Final temperature: T2=196∘C=196+273.15=469.15 KT_2 = 196^\circ C = 196 + 273.15 = 469.15 \, KT2=196∘C=196+273.15=469.15K
- Initial power radiated: P1=92 J/sP_1 = 92 \, J/sP1=92J/s
Step-by-Step Process:
The power radiated is proportional to the fourth power of the temperature:P2P1=(T2T1)4\frac{P_2}{P_1} = \left( \frac{T_2}{T_1} \right)^4P1P2=(T1T2)4
Substitute the known values:P292=(469.15385.15)4\frac{P_2}{92} = \left( \frac{469.15}{385.15} \right)^492P2=(385.15469.15)4
First, calculate the ratio of temperatures:469.15385.15≈1.219\frac{469.15}{385.15} \approx 1.219385.15469.15≈1.219
Raise this ratio to the fourth power:1.2194≈2.0731.219^4 \approx 2.0731.2194≈2.073
Thus,P2=92×2.073≈190.7 J/sP_2 = 92 \times 2.073 \approx 190.7 \, \text{J/s}P2=92×2.073≈190.7J/s
Conclusion:
The rate at which the cube will radiate heat when its surface temperature is increased to 196 °C is approximately 191 J/s.
