Calculate the amount of time required to raise temperature of 0.50 liter of water from

0.0° C to 10.0° C if heat energy is supplied at a rate of 1100 joules per second

19 seconds

Explanation:

Heat energy was supplied and used to heat 0.50 liters of water from 0°C to 10°C.

This means water gained heat energy that was supplied.

Therefore;

Heat supplied = Heat gained by water

We are required to calculate the time taken to raise the temperature of water from 0°C to 10°C.

Step 1: Calculate the heat gained by water

Quantity of heat = Mass * specific heat capacity * change in temperature

Density of water 1 g/ml

Volume of water is 500 mL

Therefore, since Mass = density * volume

The mass of water = 500 g

Change in temperature = 10° C

Specific heat capacity of water = 4.18 J/g/°C

Thus;

Quantity of heat = 500 g * 4.18 J/g/°C * 10

= 20,900 Joules

Step 2: Heat supplied by a heater

Heat supplied = Power * time

Power = 1100 J/s

Assuming the time required is x

Heat supplied = 1100x Joules

Step 3: Time required

Remember; heat supplied = heat gained by water

Therefore;

1100x joules = 20,900 Joules

x = 20,900/1100

= 19 seconds

Therefore, the time required to raise the temperature of water from 0°C to 10°C is 19 seconds.