A person weighing 50 kg takes in 1500 kcal diet per day. If this energy were to be used in heating the body of person without any losses, then the rise in his temperature is (specific heat of human body = 0.83 cal g-1 C-1)

Question

A person weighing 50 kg takes in 1500 kcal diet per day. If this energy were to be used in heating the body of person without any losses, then the rise in his temperature is (specific heat of human body = 0.83 cal g-1 C-1) (A) 30 °C (B) 48 °C (C) 40.16 °C (D) 36.14 °C

Tardigrade Admin · Accepted Answer

Here, m=50 kg=50 × 103 g Δ Q=1500 kcal=1.5 × 106 cal s=0.83 cal g-1 °C-1 As Δ Q=msΔ T Δ T=(Δ Q/ms)=(1.5 × 106/50 × 103 × 0.83) =36.14°C

Q. A person weighing $50 k g$ takes in $1500 k c a l$ diet per day. If this energy were to be used in heating the body of person without any losses, then the rise in his temperature is (specific heat of human body $= 0.83 c a l g^{- 1} C^{- 1}$ )

$30^{\circ} C$

$48^{\circ} C$

$40.16^{\circ} C$

$36.14^{\circ} C$

Here, $m = 50 k g = 50 \times 1 0^{3} g$
$Δ Q = 1500 k c a l = 1.5 \times 1 0^{6} c a l$
$s = 0.83 c a l g^{- 1}^{\circ} C^{- 1}$
As $Δ Q = m s Δ T$
$Δ T = \frac{Δ Q}{m s} = \frac{1.5 \times 1 0 ^{6}}{50 \times 1 0 ^{3} \times 0.83}$
$= 36.1 4^{\circ} C$

Q. A person weighing 50kg takes in 1500kcal diet per day. If this energy were to be used in heating the body of person without any losses, then the rise in his temperature is (specific heat of human body =0.83calg−1C−1)

30∘C

48∘C

40.16∘C

36.14∘C