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3 years ago

100 ml of water is initially at 20°C. 30,000 J of heat is added to the water. What is temperature change for the water?

Physics

1 answer:

Agata [3.3K]3 years ago

3 0

Answer:

Therefore the new temperature of the water is 91.67°C

Explanation:

Specific heat of water = 4.186 joule /gram °C

The equation to calculate change of temperature

Q=s×m×ΔT

Q = the amount of heat = 30,000 J

s= Specific heat of water = 4.186 joule /gram °C

m = mass of the water = volume ×density=(100×1)gram =100 gram

ΔT = the change of temperature = (T-20)

Therefore,

30000= 4.186×100×(T-20)

⇔(T-20)= 71.67

⇔T = 71.67+20

⇔T = 91.67

Therefore the new temperature of the water is 91.67°C

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4 years ago

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3 years ago

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Answer:

Part a)

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Part b)

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now we know that

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so we have

$\frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mr^2)(\frac{v}{r})^2 = mgh$

$\frac{1}{2}mv^2 + \frac{1}{5}mv^2 = mgh$

$\frac{7}{10}mv^2 = mgh$

$\frac{1}{2}mv^2 = \frac{10}{14}mgh$

so the height on the smooth side is given as

$h' = \frac{10}{14} h$

Part b)

if both sides are rough then it will reach the same height on the other side because the energy is being conserved.

Part c)

Since marble will go to same height when it is rough while when it is smooth then it will go to the height

$h' = \frac{10}{14} h$

so on smooth it will go to lower height

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4 years ago

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Answer:

Explanation:

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$3.61-1.2936=0.005\times k$

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Answer:

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4 years ago