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yanalaym [24]
3 years ago
7

A pan containing 20.0 grams of water was allowed to cool from a temperature of 95.0 °C. If the amount of heat released is 1,200

joules, what is the approximate final temperature of the water? 75 °C 78 °C 81 °C 87 °C
Chemistry
1 answer:
Sedbober [7]3 years ago
7 0

Answer:

81°C.

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).

∵ Q = m.c.ΔT

∴ (- 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: 81°C.</em>

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