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Chapter 5: Q8E (page 268)

Question: How much would the temperature of 275 g of water increase if 36.5 kJ of heat were added?

Short Answer

Expert verified

The rise in the temperature of water = 31.72\(^0C\)

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� T.

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 = 275 g

Q = 36.5kJ = 36500 J

By putting the values above in the equation Q = C 脳 m 脳 鈭� T, we get:

36500 = 4.184\( \times \)275\( \times \)\(\Delta \)T.

\(\Delta \)T = \(\frac{{36500}}{{4.184 \times 275}} = {31.72^0}C\).

Therefore, the rise in the temperature of 275 g of water = 31.72\(^0C\).

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Most popular questions from this chapter

The addition of 3.15g of Ba(OH)2.8H2O to a solution of 1.52g of NH4SCN in 100g of water in a calorimeter caused the temperature to fall by 3.1藲C. Assuming the specific heat of the solution and products is 4.20J/g藲C, calculate the approximate amount of heat absorbed by the reaction, which can be represented by the following equation:
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