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Maslowich
3 years ago
14

A 125-g metal block at a temperature of 93.2 °C was immersed in 100. g of water at 18.3 °C. Given the specific heat of the metal

( s = 0.900 J/g·°C), find the final temperature of the block and the water
Physics
1 answer:
Nataly_w [17]3 years ago
6 0

Answer:

34.17°C

Explanation:

Given:

mass of metal block = 125 g

initial temperature T_i = 93.2°C

We know

Q = m c \Delta T   ..................(1)

Q= Quantity of heat

m = mass of the substance

c = specific heat capacity

c = 4.19 for H₂O in J/g^{\circ}C

\Delta T = change in temperature

Now

The heat lost by metal = The heat gained by the metal

Heat lost by metal = 125\times 0.9\times (93.2-T_f)

Heat gained by the water = 100\times 4.184\times(T_f -18.3)

thus, we have

125\times 0.9\times (93.2-T_f) = 100\times 4.184\times(T_f -18.3)

10485-112.5T_f = 418.4T_f - 7656.72

⇒ T_f = 34.17^oC

Therefore, the final temperature will be = 34.17°C

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Explanation:

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3 years ago
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NISA [10]

Answer:

a)  The swimmer should travel perpendicular to the bank to minimize the spent in getting to the other side.

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c) 53.13°

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7 0
3 years ago
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ivann1987 [24]

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