Answer:
23.6°C
Explanation:
Step 1:
Data obtained from the question. This include the following:
Mass = 0.125kg = 0.125 x 1000 = 125g
Initial temperature (T1) = 22°C
Heat (Q) generated in 4.5mins = 835J
Specific heat capacity (C) of water = 4.184J/g°C
Change in temperature (ΔT) =?
Final temperature (T2) =?
Step 2:
Determination of the change in temperature, ΔT of water.
This can be obtained as shown below:
Q = MCΔT
835 = 125 x 4.184 x ΔT
Divide both side by 125 x 4.184
ΔT = 835 / (125 x 4.184)
ΔT = 1.6°C
Therefore, the change in temperature, ΔT is 1.6°C
Step 3:
Determination of the final temperature of water.
This can be obtained as follow:
Initial temperature (T1) = 22°C
Change in temperature (ΔT) = 1.6°C
Final temperature (T2) =?
Change in temperature (ΔT) = Initial temperature – Final temperature
ΔT = T2 – T1
1.6 = T2 – 22
Collect like terms
T2 = 1.6 + 22
T2 = 23.6°C
Therefore, the final temperature of the water is 23.6°C.
