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2 years ago

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ANSWER ASAP! WILL GIVE BRAINLIEST! A pan containing 20.0 grams of water was allowed to cook from a temperature of 95.0 degrees C

elsius. If the amount of heat released is 1,200 joules, what is the approximate final temperature of the water?
A. 75 degrees Celsius

B. 78 degrees Celsius

C. 81 degrees Celsius

D. 87 degrees Celsius

2 answers:

vodka [1.7K]2 years ago

3 0

Answer:

C. 81 degrees Celsius

Explanation:

To solve this problem, we can use the relation:

<em>Q = m.c.ΔT,</em>

where, Q is the amount of heat released from water (Q = - 1200 J).

m is the mass of the water (m = 20.0 g).

c is the specific heat capacity of water (c of water = 4.186 J/g.°C).

ΔT is the difference between the initial and final temperature (ΔT = final T - initial T = final T - 95.0°C).

<em>∵ Q = m.c.ΔT</em>

∴ (- 1200 J) = (20.0 g)(4.186 J/g.°C)(final T - 95.0°C ).

(- 1200 J) = 83.72 final T - 7953.

∴ final T = (- 1200 J + 7953)/83.72 = 80.67 °C ≅ 81.0 °C.

<em>So, the right choice is: C. 81 degrees Celsius </em>

<em></em>

JulijaS [17]2 years ago

3 0

Answer : The correct option is, (C) $81^oC$

Explanation :

Formula used :

$Q=m\times c\times \Delta T$

or,

$Q=m\times c\times (T_2-T_1)$

where,

Q = heat released = -1200 J

m = mass of water = 20.0 g

c = specific heat of water = $4.184J/g.^oC$

$T_1$ = initial temperature = $95.0^oC$

$T_2$ = final temperature = ?

Now put all the given value in the above formula, we get:

$-1200J=20.0g\times 4.184J/g.^oC\times (T_2-95.0)^oC$

$T_2=80.6^oC\approx 81^oC$

Therefore, the approximate final temperature of the water is $81^oC$

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