Question

The energy transferred to the water in 100 seconds was 155 000 J. specific heat capacity of water = 4200 J/kg °C

Answer 1

Answer:

0.47kg

Explanation:

1. find the change in temperature (the start point of the straight line)

100-22= 78 degrees celsius

2. use specific heat capacity equation

change in thermal energy (J) = m x J/kg C X 78

155000= m x 4200 x 78

3. rearrange to find m

155000/(4200x78)

4200x78= 327600

155000/327600= 0.4731379

2sf= 0.47kg

Answer 2

Answer:

0.37 kg

Explanation:

I'm not a professor myself, but this is how I worked it out:

using the graph, after 100 seconds, the temperature is 100 degrees Celsius.

If we now substitute everything into the specific heat capacity equation, making the mass "m", we would come up with:

4200 = 155000/(m x 100)

If we rearrange and solve for m, we get 0.37 kg.

I'm not sure if I have done this correctly, feel free to correct me.

Hope this helps!