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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 脳 m 脳鈭員,

Where 鈥淐鈥� is the specific heat of the substance, 鈥渕鈥� is the mass of the substance, and 鈥溾垎T鈥� 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 掳C(Table 5.1)

m = 485 g

Q = 14.5kJ = 14500 J

By putting the values above in the equation Q = C 脳 m 脳鈭� 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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