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A tree leaf of mass of 0.80 g and specific heat of 3700J/kgC3700 \mathrm{J} / \mathrm{kg} \cdot^{\circ} \mathrm{C} absorbs energy from the sunlight at a rate of 2.8J/s2.8 \mathrm{J} / \mathrm{s}.If this energy is not removed from the leaf,how much does the temperature of the leaf change in 1 min?

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Answered 2 years ago
Answered 2 years ago
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we have that

Q=HΔt=2.8×1×60=168 J\begin{align*} Q&=H \Delta t\\ &=2.8 \times 1 \times 60\\ &=168 \text{ J} \end{align*}

  if this energy increase the leaf temperature 

  then we have

Q=mcΔtΔt=Qmc=1680.80×103×3700=56.77oC\begin{align*} Q&=mc\Delta t\\ \Delta t&=\dfrac{Q}{m c}\\ &=\dfrac{168}{0.80 \times 10^{-3} \times 3700}\\ &=\boxed{56.77^o\text{C}} \end{align*}

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