Click to Chat

1800-2000-838

+91-120-4616500

CART 0

• 0

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
`        An icebox almost completely filled with ice at 0°C is dipped into a large volume of water at 20°C. The box has walls of surface area 2400 cm2. Thickness 2.0 mm and thermal conductivity 0.06 W/m-°C. Calculate the rate at which the ice melts in the box. Latent heat of fusion of ice = 3.4 × 105 J/kg.`
3 years ago

Kevin Nash
332 Points
```										Sol. A = 2400 cm2 = 2400 × 10^–4 m^2
ℓ = 2 mm = 2 × 10^–3 m
K = 0.06 w/m-°C
θ base 1 = 20°C
θ base 2 = 0°C
Q/t = KA(θ base 1 – base 2)/ℓ = 0.06 * 2400 * 10^-4 * 20/2 * 10^-3 = 24 * 6 * 10^-1 * 10 = 24 * 6 = 144 j/sec
Rate in which ice melts = m/t = Q/t * L = 144/3.4 * 10^5 Kg/h = 144 * 366/3.4 * 10^4 Kg/s = 1.52 kg/s.

```
3 years ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Thermal Physics

View all Questions »
• Complete Physics Course - Class 12
• OFFERED PRICE: Rs. 2,756
• View Details
Get extra Rs. 413 off
USE CODE: COUPON10
• Complete Physics Course - Class 11
• OFFERED PRICE: Rs. 2,968
• View Details
Get extra Rs. 445 off
USE CODE: COUPON10