Question

A 300.0g sample of glass at 76.0C was placed in a large calorimeter containing 1000.0g of water. The water temperature was initially 23.9C. After the heat transfer, the water temp rose to 27.0C. Calculate the specific heat of the glass; then determine the percent error in this value.

Answer 1

For a 300.0g sample of glass at 76.0C  placed in a large calorimeter containing 1000.0g of water,  the specific heat of the glass and the percent error in this value are mathematically given as

C= 0.211 cal / g

% E = 5.5%

What are the specific heat of the glass and the percent error in this value.?

Generally, the equation for the  balanced heat on glass is mathematically given as

Mg* SH of glass *dT of glass = Mwater * SH of water * dT of water

Therefore

300 * C * (76 - 27) = 1000 * 1 * (27 - 23.9)

C = 1000 * 3.1 / (300 * 49)

C= 0.211 cal / g

In conclusion,The precent error is

% E = (0.211 - 0.2) * 100 / 0.2

% E = 5.5%

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