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Artyom0805 [142]
3 years ago
15

A 23.5g aluminum block is warmed to 65.9°C and plunged into an insulated beaker containing 55.0g water initially at 22.3°C. The

aluminum and the water are allowed to come to thermal equilibrium.
Assuming that no heat is lost, what is the final temperature of the water and aluminum?
Chemistry
1 answer:
atroni [7]3 years ago
4 0

Answer:

25.97oC

Explanation:

Heat lost by aluminum = heat gained by water

M(Al) x C(Al) x [ Temp(Al) – Temp(Al+H2O) ] = M(H2O) x C(H2O) x [ Temp(Al+H2O) – Temp(H2O) ]

Where M(Al) = 23.5g, C(Al) = specific heat capacity of aluminum = 0.900J/goC, Temp(Al) = 65.9oC, Temp(Al+H2O)= temperature of water and aluminum at equilibrium = ?, M(H2O) = 55.0g, C(H2O)= specific heat capacity of liquid water = 4.186J/goC

Let Temp(Al+H2O) = X

23.5 x 0.900 x (65.9-X) = 55.0 x 4.186 x (X-22.3)

21.15(65.9-X) = 230.23(X-22.3)

1393.785 - 21.15X = 230.23X – 5134.129

230.23X + 21.15X = 1393.785 + 5134.129

251.38X = 6527.909

X = 6527.909/251.38

X = 25.97oC

So, the final temperature of the water and aluminum is = 25.97oC

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Whats the balanced equation for sodium hydroxide + ammonium sulphate=sodium sulphate + amonia + h2o​
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The answer is option C.

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7 0
4 years ago
Read 2 more answers
A piece of unknown metal with mass 30 g is heated to 110.0 °C and dropped into 100.0 g of water at 20.0 °C. The final temperatur
Ymorist [56]

<u>Answer:</u> The specific heat of metal is 0.821 J/g°C

<u>Explanation:</u>

When metal is dipped in water, the amount of heat released by metal will be equal to the amount of heat absorbed by water.

Heat_{\text{absorbed}}=Heat_{\text{released}}

The equation used to calculate heat released or absorbed follows:

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]       ......(1)

where,

q = heat absorbed or released

m_1 = mass of metal = 30 g

m_2 = mass of water = 100 g

T_{final} = final temperature = 25°C

T_1 = initial temperature of metal = 110°C

T_2 = initial temperature of water = 20.0°C

c_1 = specific heat of metal = ?

c_2 = specific heat of water = 4.186 J/g°C

Putting values in equation 1, we get:

30\times c_1\times (25-110)=-[100\times 4.186\times (25-20)]

c_1=0.821J/g^oC

Hence, the specific heat of metal is 0.821 J/g°C

8 0
3 years ago
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