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Chapter 5: Q9E (page 268)

Question: If 14.5 kJ of heat were added to 485 g of liquid water, how much would its temperature increase?

Short Answer

Expert verified

The rise in the temperature of water = \({7.14^0}C\).

Step by step solution

01

Specific heat

The heat required to raise the temperature of a substance is given by the formula Q = C × m ×∆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

Increase in temperature

We know from the given details that:

C = 4.184 J/g °C(Table 5.1)

m = 485 g

Q = 14.5kJ = 14500 J

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

14500 = 4.184 \( \times \) 485 \( \times \)\(\Delta \)T.

\(\Delta \)T = \(\frac{{14500}}{{4.184 \times 485}} = {7.14^0}C\).

Therefore, the rise in the temperature of 485 g of water = \({7.14^0}C\).

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Most popular questions from this chapter

How many milliliters of water at 23 °C with a density of 1.00 g/mL must be mixed with 180 mL (about 6 oz) of coffee at 95 °C so that the resulting combination will have a temperature of 60 °C? Assume that coffee and water have the same density and the same specific heat.

Question: Prepare a table identifying the several energy transitions that take place during a typical operation of an automobile.

Question 11: A piece of unknown solid substance weighs 437.2 g, and requires 8460 J to increase its temperature from 19.3 °C to 68.9 °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?

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