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Classical Mechanics Level 3

A person's body converts chemical energy into internal energy at a metabolic rate of 100 kcal/h 100\text{ kcal/h} . The person stands neck-deep in an insulated tub containing 0.8 m 3 0.8\text{ m}^3 of water at 2 0 ∘ C 20^\circ C . Disregarding energy loss from the tub, determine the change of water temperature in 45 minutes.

Specific Heat of water is 1 cal/g ∘ C 1 \text{ cal/g}^\circ C .

(Answer In ∘ C ^\circ C )

95.75 93.75 98.9 96.35

Kushagra Yadav by Kushagra Yadav , 24, India

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2 solutions

Sravanth C.
Jan 29, 2016

The rate at which the energy is being converted = 100 kcal/h =100\text{kcal/h} . Therefore in 45 45 min the amount of energy lost will be 100 kcal/h × 45 60 = 75 kcal = Q \dfrac{100\text{kcal/h}\times 45}{60}= 75\text{kcal}=Q .

And the volume of water = 0.8 m 3 = 800 L = 800 k g = m =0.8m^3=800L=800kg=m .

We know that Q = C m Δ T Q=Cm\Delta T . Therefore, Δ T = Q C m Δ T = 75000 1 × 800 = 93.75 \Delta T= \dfrac{Q}{Cm}\\\Delta T = \dfrac{75000}{1\times 800}=\boxed{93.75}

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Kaustubh Miglani
Mar 1, 2016

@Kushagra Yadav Whats ur score in JSTSE by REVISED ANSWER KEY on edudel?

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