## 4A.13

Shimei_2F
Posts: 100
Joined: Fri Aug 09, 2019 12:17 am
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### 4A.13

Problem: A constant-volume calorimeter was calibrated by carrying out a reaction known to release 3.50 kJ of heat in 0.200 L of solution in the calorimeter (q 5 23.50 kJ), resulting in a temperature rise of 7.32 C. In a subsequent experiment, 100.0 mL of 0.200 m HBr(aq) and 100.0 mL of 0.200 m KOH(aq) were mixed in the same calorimeter and the temperature rose by 2.49 C. What is the change in the internal energy of the reaction mixture as a result of the neutralization reaction?

When we calculate the heat capacity of the calorimeter, why do we change the unit of temperature from celsius to kelvin without adding 273 to 7.32 C?

Kate Swertfager
Posts: 50
Joined: Wed Sep 18, 2019 12:17 am

### Re: 4A.13

I think because the temp change between temp in K and temp in C is the same?

Selena Yu 1H
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### Re: 4A.13

Yes it is because a change in temperature is the same no matter the units and so a change in temperature of 7.32 C is the same as a change in temperature of 7.32 K

nicolely2F
Posts: 149
Joined: Sat Sep 14, 2019 12:17 am

### Re: 4A.13

Kate Swertfager wrote:I think because the temp change between temp in K and temp in C is the same?

Correct. The units are different but when it comes to changes in temperature, the change value in both Celsius and Kelvin is numerically the same.