If 2.0 g of water at 0.00°C is to be vaporized, how much heat must be added to it? The specific heat of water is 1.0 cal/g?K, its heat of fusion is 80 cal/g, and its heat of vaporization is 539 cal/g

A) 1100 kcal B) 1200 cal C) 1500 cal D) 1100 cal E) 1300 cal


E

Physics & Space Science

You might also like to view...

Which terrestrial planet has the oldest surface?

A. Venus B. Earth C. Mercury D. Mars

Physics & Space Science

According to quantum field theory, the universe if made of

A) photons. B) quantized fields. C) electromagnetic fields. D) baseball fields. E) particles moving in empty space.

Physics & Space Science

A light oil flows through a copper tube of 2.6-cm-ID and 3.2-cm-OD. Air flows perpendicular over the exterior of the tube as shown in the following sketch. The convection heat transfer coefficient for the oil is 120 W/(m2 K) and for the air is 35 W/(m2 K). Calculate the overall heat transfer coefficient based on the outside area of the tube (a) considering the thermal resistance of the tube, (b) neglecting the resistance of the tube.

GIVEN
• Air flow over a copper tube with oil flow within the tube
• Tube diameters
? Inside (Di) = 2.6 cm = 0.026 m
? Outside (Do) = 3.2 cm = 0.032 m
• Convective heat transfer coefficients
? Oil h i= 120 W/(m2 K)
? Air h o= 35 W/(m2 K)
FIND
• The overall heat transfer coefficient based on the outside tube area (Uo), (a) considering the thermal resistance of the tube, and (b) neglecting the tube resistance
ASSUMPTIONS
• Uniform heat transfer coefficients
• Variation of thermal properties is negligible
SKETCH

PROPERTIES AND CONSTANTS
the thermal conductivity of copper (k) = 392 W/(m K) (at 127°C)

Physics & Space Science

The north celestial pole appears 30 degrees above your horizon. The star Vega is on your meridian. By studying your star charts and your clocks, you determine that Vega crossed the meridian in Greenwich (England) 3 hours ago. Where are you?

A) latitude 30°, longitude 45°W B) latitude 30°N, longitude 3°W C) latitude 60°N, longitude 45°W D) It is not possible to determine your longitude from the information given.

Physics & Space Science