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skelet666 [1.2K]
3 years ago
6

If the specific heat of water is 1.0 calorie/gram°C, how many calories are required to raise 500 grams of water 10.0°C?

Physics
1 answer:
kykrilka [37]3 years ago
6 0
The amount of heat that is required to raise the temperature of a system is calculated by the expression:

Heat required = m Cp ΔT
Heat required = 500 g (1.0 calorie/gram °C) (10.0 <span>°C)
Heat required = 5000 calorie
 
The correct answer from the choices listed above is the second option.</span>
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Order the speed of sound through these materials from the slowest to the fastest.
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Speed of sound in cold air < Speed of sound in Warm air < Speed of sound in hot molten lead < Speed of sound in water

Explanation:

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Speed of sound in water varies from 1450 to 1498 meters per second

Speed of sound in Hot Molten lead is approximately 1210 meters per second

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Speed of sound in cold air is approximately 293.33 meters per second

Step 2:

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8 0
3 years ago
. An object whose mass is 375 lb falls freely under the influence of gravity from an initial elevation of 253 ft above the surfa
lozanna [386]

Answer:

(a) Vf = 128 ft/s

(b) K.E = 122.8 Btu

Explanation:

(a)

In order to find the velocity of the object just before striking the surface of earth or the final velocity, we use 3rd equation of motion:

2gh = Vf² - Vi²

where,

g = 32.2 ft/s²

h = height = 253 ft

Vf = Final Velocity = ?

Vi = Initial Velocity = 10 ft/s

Therefore,

(2)(32.2 ft/s²)(253 ft) = Vf² - (10 ft/s)²

16293.2 ft²/s² + 100 ft²/s² = Vf²

Vf = √(16393.2 ft²/s²)

<u>Vf = 128 ft/s</u>

<u></u>

(b)

The kinetic energy of the object before it hits the surface of earth is given by:

K.E = (0.5)(m)(Vf)²

where,

m = mass of object = 375 lb

K.E = Kinetic energy of object before it strikes the surface of earth = ?

Therefore,

K.E = (0.5)(375 lb)(128 ft/s)²

K.E = 3073725 lb.ft²/s²

Now, converting this to Btu:

K.E = (3073725 lb.ft²/s²)(1 Btu/25037 lb.ft²/s²)

<u>K.E = 122.8 Btu</u>

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4 years ago
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