Heat required to melt 1 g of ice is 80 cal. A man melts 60 g of ice by chewing in one minute. His power is:

Question

Heat required to melt 1 g of ice is 80 cal. A man melts 60 g of ice by chewing in one minute. His power is: (A) 4800 W (B) 336 W (C) 1.33 W (D) 0.75 W

Tardigrade Admin · Accepted Answer

= ML = 60×80 = 4800cal = 4800 × 4.2J. P = (Q/T)=(4800×4.2/60)=336 W

Q. Heat required to melt $1 g$ of ice is $80 c a l$ . A man melts $60 g$ of ice by chewing in one minute. His power is:

4800 W

336 W

1.33 W

0.75 W

= ML = $60 \times 80$ = $4800 c a l$ = $4800 \times 4.2 J$ .
P = $\frac{Q}{T} = \frac{4800 \times 4.2}{60} = 336 W$

Q. Heat required to melt 1g of ice is 80cal. A man melts 60g of ice by chewing in one minute. His power is: