A normal diet furnishes 200 kcal to a 60 kg person in a day. If this energy was used to heat the person with no losses to the surroundings, how much would the person's temperature increases? The specific heat of the human body is 0.83 cal g -1 ° C -1.

Question

A normal diet furnishes 200 kcal to a 60 kg person in a day. If this energy was used to heat the person with no losses to the surroundings, how much would the person's temperature increases? The specific heat of the human body is 0.83 cal g -1 ° C -1. (A) 8.2° C (B) 4.01° C (C) 6.0° C (D) 5.03° C

Tardigrade Admin · Accepted Answer

Given, m=60 kg =60 × 103 g , s=0.83 cal g -1 ° C -1 Q=200 kcal =2 × 105 cal Amount of heat required for a person, Q=m s Δ T ⇒ Δ T =(Q/m s)=(2 × 105/60 × 103 × 0.83) =4.01° C So, the person's temperature increases by 4.01° C.

Q. A normal diet furnishes $200 k c a l$ to a $60 k g$ person in a day. If this energy was used to heat the person with no losses to the surroundings, how much would the person's temperature increases? The specific heat of the human body is $0.83 c a l g^{- 1}^{\circ} C^{- 1}$ .

$8. 2^{\circ} C$

$4.0 1^{\circ} C$

$6. 0^{\circ} C$

$5.0 3^{\circ} C$

Q. A normal diet furnishes 200kcal to a 60kg person in a day. If this energy was used to heat the person with no losses to the surroundings, how much would the person's temperature increases? The specific heat of the human body is 0.83calg−1∘C−1.

8.2∘C

4.01∘C

6.0∘C

5.03∘C

Given, m=60kg=60×103g,s=0.83calg−1∘C−1 Q=200kcal=2×105cal Amount of heat required for a person, Q=msΔT ⇒ΔT=msQ​=60×103×0.832×105​ =4.01∘C So, the person's temperature increases by 4.01∘C.

Q. A normal diet furnishes $200 k c a l$ to a $60 k g$ person in a day. If this energy was used to heat the person with no losses to the surroundings, how much would the person's temperature increases? The specific heat of the human body is $0.83 c a l g^{- 1}^{\circ} C^{- 1}$ .

$8. 2^{\circ} C$

$4.0 1^{\circ} C$

$6. 0^{\circ} C$

$5.0 3^{\circ} C$