Respuesta :
42.34 g of water could be warmed from 21.4°C to 43.4°C  by the pellet dropped inside it
Heat loss by the pellet is equal to the Heat gained by the water.
[tex]q_{w} = -q_{p}[/tex] ….(1)
where, [tex]q_{w}[/tex] is the heat gained by water
[tex]q_{p}[/tex] is the heat loss by pellet
[tex]q_{w}[/tex] = mCΔT
where m = mass of water
C = specific heat capacity of water = 4.184 J/g-°C
ΔT = Increase in temperature
ΔT for water = 43.4 - 21.4 = 22°C
[tex]q_{w}[/tex] = m × 4.184 × 22 …. (2)
Now
[tex]q_{p}[/tex] = [tex]H_{c}[/tex] ×ΔT
where [tex]H_{c}[/tex] = Heat capacity of pellet = 56J/°C
Δ T for pellet = 43.4 - 113 =- 69.6°C
[tex]q_{p}[/tex] = 56 × -69.6 = -3897.6 J
From equation (1) and (2)
-m× 4.184 × 22 =-3897.6
m= 42.34 g
Hence, 42.34 g of water could be warmed from 21.4 degrees Celsius to 43.4 degrees Celsius by the pellet dropped inside it.
Learn more about specific heat here https://brainly.com/question/16559442
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