Solved: Use the heat equation to calculate the energy, in joules and calories, for each

Chapter 0, Problem 3.37

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QUESTION:

Use the heat equation to calculate the energy, in joules and calories, for each of the following (see Table 3.11):

a. to heat 25.0 g of water from 12.5 °C to 25.7 °C

b. to heat 38.0 g of copper from 122 °C to 246 °C

c. lost when 15.0 g of ethanol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\), cools from 60.5 °C to -42.0 °C

d. lost when 125 g of iron cools from 118 °C to 55 °C

Questions & Answers

QUESTION:

Use the heat equation to calculate the energy, in joules and calories, for each of the following (see Table 3.11):

a. to heat 25.0 g of water from 12.5 °C to 25.7 °C

b. to heat 38.0 g of copper from 122 °C to 246 °C

c. lost when 15.0 g of ethanol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\), cools from 60.5 °C to -42.0 °C

d. lost when 125 g of iron cools from 118 °C to 55 °C

ANSWER:

Step 1 of 5

The physical property termed as specific heat is used to explain about the energy requirements of several substances. The quantity of heat needed to elevate the temperature of 1 g of a specific substance is referred to as the specific heat (SH).

The heat lost or gained by a particular substance can be obtained using the formula:

                                                          

Where,

m denotes the mass of the substance.

C denotes the specific heat capacity.

  denotes the change in temperature.

 

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