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32.0°C to 61.0°C, what is the specific heat of the substance? Use q=m.cp.ΔT

a. 0.171 J/(g°C)
b. 0.548 J/(g°C)
c. 15.9 J/(g°C)
d. 86.6 J/(g°C)

1 Answer

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Final answer:

The specific heat (c) of the substance is calculated using the formula c = q / (m.ΔT). For the given mass of 437.2 g, heat of 8460 J, and a temperature change of 49.6 °C, the specific heat is found to be 0.390 J/(g°C).

Step-by-step explanation:

To calculate the specific heat (c) of a substance using the heat equation (q = m.cp.ΔT), you will need to know the mass of the substance, the amount of heat added or removed, and the temperature change that the substance undergoes. The specific heat is an intensive property and represents the amount of heat required to raise the temperature of 1 gram of the substance by 1 degree Celsius. You can rearrange the formula to solve for c as c = q / (m.ΔT). With the provided information, the mass of the substance is 437.2 g, the heat added is 8460 J, and the temperature change (ΔT) is (68.9 °C - 19.3 °C) = 49.6 °C. Plugging these values into the equation gives you c = 8460 J / (437.2 g * 49.6 °C) which equals 0.390 J/(g°C), matching with the provided option (a).

User Dmitriy Dumanskiy
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