JEE Challenger
More from Thermal Properties of Matter
PhyMedAdv 2020 • Paper 2

Temperature Difference of Water Under Sunlight Using Newtons Law of Cooling

A container holding 1 kg1\text{ kg} of water is exposed to solar radiation, heating it above ambient temperature. The solar radiation provides an average power per unit area of 700 W m−2700\text{ W m}^{-2} absorbed over an effective surface area of 0.05 m20.05\text{ m}^2. Assuming heat loss is governed by Newton's law of cooling, calculate the temperature difference (in ∘C^{\circ}\text{C}) between the water and surroundings after a long time. (Ignore the heat capacity of the container; take Newton's cooling constant as 0.001 s−10.001\text{ s}^{-1} and heat capacity of water as 4200 J kg−1 K−14200\text{ J kg}^{-1}\text{ K}^{-1}).

Official Numerical Answer8.33

Topics & Concepts

Thermal Properties of MatterHeat TransferCalorimetry

Step-by-Step Solution

To find the temperature difference between the water and its surroundings after a long time, we analyze the thermal energy balance of the water at steady state.

1. Given Data:

  • Mass of water, m=1 kgm = 1\text{ kg}
  • Solar radiation power per unit area, I=700 W m−2I = 700\text{ W m}^{-2}
  • Effective absorption area, A=0.05 m2A = 0.05\text{ m}^2
  • Constant for Newton's law of cooling, k=0.001 s−1k = 0.001\text{ s}^{-1}
  • Specific heat capacity of water, s=4200 J kg−1 K−1s = 4200\text{ J kg}^{-1}\text{ K}^{-1}

2. Rate of Heat Energy Received:

The heat energy absorbed by the water per unit time (power input) from the sunlight is given by: Pin=I×AP_{\text{in}} = I \times A

Substituting the given values: Pin=700 W m−2×0.05 m2=35 W=35 J s−1P_{\text{in}} = 700\text{ W m}^{-2} \times 0.05\text{ m}^2 = 35\text{ W} = 35\text{ J s}^{-1}

3. Rate of Heat Loss via Newton's Law of Cooling:

According to Newton's law of cooling, the rate of temperature change of a body is proportional to the temperature difference between the body and its surroundings: (dTdt)loss=k(T−T0)\left(\frac{dT}{dt}\right)_{\text{loss}} = k (T - T_0)

where:

  • TT is the temperature of the water,
  • T0T_0 is the ambient temperature of the surroundings,
  • (T−T0)=ΔT(T - T_0) = \Delta T is the temperature difference.

The corresponding rate of heat loss to the surroundings is: dQlossdt=ms(dTdt)loss=msk(T−T0)\frac{dQ_{\text{loss}}}{dt} = m s \left(\frac{dT}{dt}\right)_{\text{loss}} = m s k (T - T_0)

4. Steady-State Analysis (After a Long Time):

After a long time, the temperature of the water reaches a constant steady-state value, meaning the net rate of heat transfer is zero: dQnetdt=Pin−dQlossdt=0\frac{dQ_{\text{net}}}{dt} = P_{\text{in}} - \frac{dQ_{\text{loss}}}{dt} = 0

  ⟹  Pin=dQlossdt\implies P_{\text{in}} = \frac{dQ_{\text{loss}}}{dt}

Substituting the expression for dQlossdt\frac{dQ_{\text{loss}}}{dt}: Pin=mskΔTP_{\text{in}} = m s k \Delta T

Solving for the temperature difference ΔT\Delta T: ΔT=Pinmsk\Delta T = \frac{P_{\text{in}}}{m s k}

5. Calculation:

Substitute the known parameters: ΔT=351×4200×0.001\Delta T = \frac{35}{1 \times 4200 \times 0.001}

ΔT=354.2=35042=253∘C≈8.33∘C\Delta T = \frac{35}{4.2} = \frac{350}{42} = \frac{25}{3}^{\circ}\text{C} \approx 8.33^{\circ}\text{C}

Final Answer:

The difference in the temperature of water and the surroundings after a long time will be 8.33∘C8.33^{\circ}\text{C} (or 253∘C\frac{25}{3}^{\circ}\text{C}).

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