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Answer :
Answer: The heat absorbed by the water is 52.823 J.
Explanation:
Given: Mass of metal = 5.05 g
Specific heat of water = 4.184 [tex]J/g^{o}C[/tex]
Initial temperature = [tex]24.8^{o}C[/tex]
Final temperature = [tex]27.3^{o}C[/tex]
Formula used to calculate heat absorbed is as follows.
[tex]q = m \times C \times (T_{2} - T_{1})[/tex]
where,
q = heat
m = mass of substance
[tex]T_{1}[/tex] = initial temperature
[tex]T_{2}[/tex] = final temperature
Substitute the values into above formula as follows.
[tex]q = m \times C \times (T_{2} - T_{1})\\= 5.05 g \times 4.184 J/g^{o}C \times (27.3 - 24.8)^{o}C\\= 52.823 J[/tex]
Thus, we can conclude that heat absorbed by the water is 52.823 J.
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