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sattari [20]
3 years ago
9

You have to heat up a 1.000 gram ingot of aluminum from initial temperature Ti = 847 K to its melting point, 933 K. Calculate th

e number of calories required for this heat up process. Type in the numeric part of your answer to the nearest 0.1 calorie. E.g., if your answer is 31.38 calories, than type 31.4 in the answer box.
Physics
1 answer:
alexdok [17]3 years ago
3 0

Answer: 18.9 calories

Explanation:

The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.

Q=m\times c\times \Delta T

Q = Heat absorbed

m= mass of substance = 1.0 g  = 0.001 kg

c = specific heat capacity  of aluminium = 921.096J/kgK

Initial temperature = T_i = 847 K

Final temperature = T_f  = 933 K

Change in temperature ,\Delta T=T_f-T_i=(933-847)K=86K

Putting in the values, we get:

Q=0.001kg\times 921.096J/kgK\times 86K

Q=79.2J=18.9cal      (1J=0.24cal)

The number of calories required for this heat up process is 18.9 calories.

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3 years ago
Can some answer 7 a and b please
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Answer:

a.After 15 second Mr Comer's speed =  1.7 m/s

b.Distance travelled by Mr.Comer in 15 seconds  =   18.75\ m

Explanation:

a. Lets recall our first equation of motion V_{f} =V_{i} + at

Now we know that V_{i} = 0.8m/s , a = 0.06\ m/s^2 and

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V_{f} =V_{i} + at

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V_{f} =0.8+ 0.9

V_{f} =1.7 m/s

Then Mr.Comer's speed after 15 sec = 1.7ms^-1

b.

Lets find the distance and recall our third equation of motion.

V_{f} ^2-V{i}^2 = 2as

So s = distance covered.

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\frac{V_{f} ^2-V{i}^2}{2a} = 2as

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4 0
3 years ago
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