MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
Menu
Grade: 12th pass
        
The question is on the imag
QUESTION 1 (10 marks)
Calculate the rate at which body heat is conducted through the clothing of a skier in a steady-state process, given the following data: the body surface area is 1.8 m2, and the clothing is 1.0 cm thick; the skin surface temperature is 33􏰓 °C and the outer surface of the clothing is at 1.0 °C; the thermal conductivity of the clothing is 0.040 W/m.K
 
QUESTION 2 (15 marks)
A loaf of freshly baked bread is left to cool on a cooling rack. The dimension of the loaf is as described in the Figure 1, and its surface temperature is 120 °C. The temperature of the air is 20 °C, and the convection coefficient from the loaf to air is 10 W/m2K. Emissivity of the bread is 0.76, and its conductivity is 0.121.
Text Box: Calculate the total heat loss from the bread. loafFigure1: Loaf of Bread
 
QUESTION 3 (10 marks)
An engine transfers 20.0 × 103 J of energy from a hot reservoir during a cycle and transfers 15.0 × 103 J as exhaust to a cold reservoir. Calculate the:
  1. amount of work done in one cycle                                                     (4 marks)
  2. efficiency of the engine                                                                      (6 marks)
QUESTION 4 (5 marks)
Explain why the entropy of the universe increases in all real processes?
e
28 days ago

Answers : (1)

Arun
15947 Points
							
Dear student
 
Please ask only one question in one thread.
 
1.)
 
 ) (305 K) / (0.012 m)= 1830 WFrom the above observation we conclude that, the rate H at which heat flows out through the clothing of a skier would be 1830 W.2 for the area A, 305 K for temperature difference ΔT and 0.012 m for thickness Δx and 0.040 W/m. K for the thermal conductivity k in the equation H = kA ΔT/ Δx,H = kA ΔT/ Δx= (0.040 W/m. K) (1.8 m2Temperature difference, ΔT = (skin surface temperature) – (outer surface temperature of clothing)= 33̊ C – 1̊ C= 32̊ C= (32+273) K= 305 KThickness, Δx = 1.2 cm= (1.2 cm) (1 m/100 cm)= 0.012 mTo find out the rate H at which heat flows out through the clothing of a skier, substitute 1.8 m2Surface area of the body,A = 1.8 m
28 days ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies

Get Extra Rs. 276 off

COUPON CODE: SELF10


Course Features

  • 101 Video Lectures
  • Revision Notes
  • Previous Year Papers
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Test paper with Video Solution


Course Features

  • 110 Video Lectures
  • Revision Notes
  • Test paper with Video Solution
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Previous Year Exam Questions


Get Extra Rs. 297 off

COUPON CODE: SELF10

Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!! Click Here for details