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Chapter 5: Q10E (page 268)

Question 10: A piece of an unknown substance weighs 44.7 g and requires 2110 J to increase its temperature from 23.2 °C to 89.6 °C.

(a) What is the specific heat of the substance?

(b) If it is one of the substances found in Table 5.1, what is its likely identity?

Short Answer

Expert verified
  1. The specific heat of the substance is 0.710\(\frac{J}{{{g^0}C}}\).
  2. The substance is most likely to be silicon (Si).

Step by step solution

01

Step 1: Specific heat of the substance

The heat required to raise the temperature of a substance is given by the formula

Q = C\( \times \)m\( \times \)\(\Delta \) T,where “C” is the specific heat of the substance, “m” is the mass of the substance, and “∆T” is the change in the temperature of the substance.

02

Calculation of specific heat

We know from the given details that:

\(\Delta \)T = \(\)\({T_{final}} - {T_{initial}} = {89.6^0}C - {23.2^0}C = {66.4^0}C\)

m = 44.7 g

Q = 2110 J

By putting the values above in the equationQ = C × m × ∆ T, we get:

2110 = 44.7\( \times \)66.4\( \times \)C.

\(\begin{array}{c}C = \frac{{2110}}{{44.7 \times 66.4}}\\ = 0.710\frac{J}{{{g^0}C}}\end{array}\)

Thus, the specific heat of the substance is 0.710\(\frac{J}{{{g^0}C}}\).

03

Identity of the substance

The specific heat of a substance is a constant value. If the value of specific heat gets matched with the table 5.1, it most likely resembles silicon (Si).

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Most popular questions from this chapter

The white pigment \({\bf{Ti}}{{\bf{O}}_{\bf{2}}}\)is prepared by the reaction of titanium tetrachloride, \({\bf{TiC}}{{\bf{l}}_{\bf{4}}}\), with water vapor in the gas phase:

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