Answer :

Sure! To calculate the rise in temperature when 84 KJ of heat is supplied to 2 kg of water, we can use the formula related to specific heat capacity:

[tex]\[ Q = m \times c \times \Delta T \][/tex]

where:
- [tex]\( Q \)[/tex] is the heat energy supplied (in joules),
- [tex]\( m \)[/tex] is the mass of the substance (in grams),
- [tex]\( c \)[/tex] is the specific heat capacity (in joules per gram per degree Celsius),
- [tex]\( \Delta T \)[/tex] is the change in temperature (in degrees Celsius).

Let's break it down step-by-step:

1. Convert the mass of water into grams:
- We have 2 kg of water. Since 1 kg = 1000 grams, the mass of water in grams is:
[tex]\[ m = 2 \, \text{kg} \times 1000 \, \text{g/kg} = 2000 \, \text{g} \][/tex]

2. Convert the heat supplied from kilojoules to joules:
- We have 84 KJ of heat. Since 1 KJ = 1000 joules, the heat supplied in joules is:
[tex]\[ Q = 84 \, \text{KJ} \times 1000 \, \text{J/KJ} = 84000 \, \text{J} \][/tex]

3. Use the specific heat capacity for water:
- The specific heat capacity of water is approximately 4.186 J/g°C.

4. Calculate the change in temperature ([tex]\( \Delta T \)[/tex]):
- Rearrange the formula to solve for [tex]\( \Delta T \)[/tex]:
[tex]\[ \Delta T = \frac{Q}{m \times c} \][/tex]
- Substitute the known values:
[tex]\[ \Delta T = \frac{84000 \, \text{J}}{2000 \, \text{g} \times 4.186 \, \text{J/g°C}} \][/tex]

5. Calculate:
- The rise in temperature of the water is approximately:
[tex]\[ \Delta T \approx 10.03 \, \text{°C} \][/tex]

So, when 84 KJ of heat is supplied to 2 kg of water, the temperature of the water increases by approximately 10.03°C.

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Rewritten by : Barada