Question

Help please, last ride guys

Answer 1

Answer:

Solution ( for fourth attachment ) : 38°C

Tip : Remember the units °C when submitting answer

Explanation:

As you mentioned, we only need the solution for the fourth attachment.

The idea here is that the heat lost by the metal will be equal to the heat gained by the water. We know that the specific heat gained or lost will always be represented by the following formula,

q = m $*$ c

Therefore if we substitute the know values and equate the two equations knowing that " q " is common among them --- ( 1 )

0.33 $*$ 448

Remember that the change in temperature of iron (ΔT) would be represented by final temperature - initial temperature, or final temperature - 693. Similarly the change in temperature of water will be final temperature - 39. Now we can pose the final temperature as a, and solve for a through substitution --- ( 2 )

0.33 $*$ 448

From here on take a look at the attachment. It represents how to receive get a through simple algebra. Here a, the final temperature, is about 38°C. In exact terms it will be $38.03617\dots$°C.

Answer 2

1. Equilibrium temperature = 7.3 C. Change in temperature of water = 2.4 C. Mass of water = 2.5 kg.
Hence energy absorbed = 4186*2.4/2.5 = 25116 J
Energy absorbed = Energy lost by brass.
Change in temperature of brass = 90.1 C. Mass = 0.52 kg. Energy lost = 25116 J
Hence specific heat = 25116/0.52/90.1 = 536 J/kgC

2. Change in temperature = 14 C
Mass of water = 195 g
Heat required to raise 1g water by 1C = 4.18 J
Hence heat required = 4.18*195*14 = 11411.4 J

For the fifth question we need the diagram or experiment above the question for context