Question

A glass containing 200.0 g of H2O at 20.0°C was placed in a refrigerator. The water loses 11.7 kJ as it cools to a constant temperature. What is its new temperature? The specific heat of water is 4.184 J/g·°C.

Answer 1

Final temperature = 6.00 °C

Step-by-step explanation:

The expression for the calculation of the enthalpy change of a process is shown below as:-

`\Delta H=m* C* \Delta T`

Where,

`\Delta H` is the enthalpy change

m is the mass

C is the specific heat capacity

`\Delta T` is the temperature change

Thus, given that:-

Mass of water = 200.0 g

Specific heat = 4.18 J/g°C

Let final temperature be x °C

`\Delta T=x-20.0\ ^0C`

`\Delta H` = -11.7 kJ (Negative sign as heat is lost)

Also, 1 kJ = 1000 kJ

`\Delta H` = -11700 J

So,

`-11700=200.0* 4.18* (x-20.0)`

`x=(1255)/(209)\\\\x=6.00\ ^0C`

Final temperature = 6.00 °C