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Question: How much heat, in joules and in calories, is required to heat a 28.4-g (1-oz) ice cube from 鈭23.0 掳C to 鈭1.0 掳C?

Short Answer

Expert verified

The heat required to raise the temperature in joules =130.77 J

The heat required to raise the temperature in calories =31.24 calories

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

Calculation of heat in joule

We know from the given details that:

C = 2.093 J/g 掳C(Table 5.1)

m = 28.4 g

\(\Delta \)T = change in temperature =\({T_{final}} - {T_{initial}} = \) -1\(^\circ C\) 鈥 ( - 23\(^0C\) ) = 22\(^0C\)

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

Q = 2.093\( \times \)28.4\( \times \)22 = 130.77 J.

The heat required to raise the temperature of 28.4 g of ice cube = 130.77 J

03

Calculation of heat in calories

Also,

1 calorie = 4.184 J.

The heat required to raise the temperature in calories =\(\frac{{130.77}}{{4.186}} = 31.24cal\)

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