Physics, asked by Kartik3813, 1 year ago

A copper calorimeter of mass 300g contains 500g of water at temperature 15 degree celsius. a 560 g block of aluminium at temperature 100 degree celsius is dropped in the calorimeter and the temperature is observed to increase to 22.5 degree celsius. find the specific heat capacity of aluminium.

Answers

Answered by gaganprasai977
32

Answer:

383.12 joule per kg degree celsius

Explanation:

Solution,

Mass of copper calorimeter (M1) = 300g = 0.3kg

Specific heat capacity of water (S2) = 4200 joule per kg degree kelivin

Mass of aluminium block(M) = 560g = 0.56 kg

Mass of water(Mw) = 0.5 kg

Temperature of water + calorimeter (k1)=15 degree celsius

Specific heat capacity of calorimeter(S1) = 390 joule per kg degree celsius

Temperature of aluminium block (k2) = 100 degree celsius

Temperature of mixture (k)= 22.5 degree celsius

Now,

Total heat lost by aluminum block

=Ms(k2-k)

=0.56*s(100-22.5)

=43.4s

we know,

heat gained by calorie

=M1*S1(k-k1)

=0.3*390(22.5-15)

=877.5

Heat gained by water

=M2S2(k-k1)

=0.5*4200*(22.5-15)

=15750

Total Heat Gained

=15750+877.5

=16627.5

Now,

Heat gained = heat lost

43.4s=16627.5

s=383.12 joule per kg degree celsius.

Answered by NehaKari
0

Answer:

The specific heat capacity of a substance can be determined by measuring the amount of heat required to raise the temperature of a given mass of the substance by a certain degree. To determine the specific heat capacity of aluminum, we can use the equation:

Q = mcΔT

where Q is the heat energy, m is the mass of aluminum, c is the specific heat capacity, and ΔT is the change in temperature.

By knowing the mass of aluminum (560 g) and the change in temperature (100 - 22.5 = 77.5 °C), we can determine the heat energy absorbed by aluminum:

Q = 560 g * 77.5 °C = 43,500 J

The heat energy absorbed by aluminum must have caused an increase in the temperature of the water in the calorimeter. We can calculate the heat energy absorbed by water using the equation:

Q = mcΔT

where m is the mass of water (500 g) and ΔT is the change in temperature (22.5 - 15 = 7.5 °C).

Q = 500 g * 4.18 J/g°C * 7.5 °C = 15,135 J

Finally, we can use the total heat energy absorbed by the water to determine the specific heat capacity of aluminum:

c = Q / (mΔT) = 15,135 J / (560 g * 77.5 °C) = 0.902 J/g°C

The specific heat capacity of aluminum is approximately 0.902 J/g°C.

#SPJ3

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