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A metal object with mass of 22.7 g is heated to 97.0 ∘C and then transferred to an insulated container containing 84.7 g of water at 20.5 ∘C. The water temperature rises and the temperature of the metal object falls until they both reach the same final temperature of 24.3 ∘C. What is the specific heat of this metal object? Assume that all the heat lost by the metal object is absorbed by the water.

Sagot :

samueladesida43

The specific heat of the metal object with a mass of 22.7g heated to to temperature of 97.0°C and then transferred to an insulated container containing 84.7 g of water at 20.5 ∘C is 0.815J/g°C

How to calculate specific heat?

The specific heat capacity of a metal can be calculated using the calorimetry equation as follows:

Q = mc∆T

Where;

Q = quantity of heat absorbed

m = mass of substance

c = specific heat capacity

∆T = change in temperature

mc∆T (water) = -mc∆T (metal)

84.7 × 4.18 × 3.8 = - (22.7 × c × -72.7)

1345.375 = 1650.29c

c = 0.815J/g°C

Therefore, the specific heat of the metal object with a mass of 22.7g heated to to temperature of 97.0°C and then transferred to an insulated container containing 84.7 g of water at 20.5 ∘C is 0.815J/g°C.

Learn more about specific heat capacity at:

#SPJ1

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