Answer:
heat transfer is when you transfer heat from one thing to another and heat transformation is when heat is transforming into something else hope it helps
Explanation:
Answer:
41.3 °C
Explanation:
From the question given above, the following data were obtained:
Mass (M) of water = 27.56 g
Heat (Q) loss = 2443 J
Final temperature (T2) = 62.5 °C
Initial temperature (T1) =?
NOTE: The specific heat capacity (C) of water is 4.18 J/g°C
Thus, we can obtain the initial temperature of the water by using the following formula:
Q = MC(T2 – T1)
2443 = 27.56 × 4.18 (62.5 – T1)
2443 = 115.2008 (62.5 – T1)
Divide both side by 115.2008
2443 / 115.2008 = (62.5 – T1)
21.20645 = 62.5 – T1
Collect like terms
21.20645 – 62.5 = – T1
– 41.3 = – T1
Divide both side by – 1
– 41.3 /– 1= – T1 / –1
41.3 = T1
T1 = 41.3 °C
Thus, the initial temperature of the water was 41.3 °C
A picture of the electroplating apparatus can be found attached. The nail (cathode) is completely submerged and that is where reduction happens. In the other side, the copper strip is (anode) where oxidation happens. The electron flow happens from the anode to the cathode. The positive charge of the battery is attached to the anode while the negative side is attached to the cathode.
