Respuesta :
Specific heat capacity is the amount of energy that required to increase the temperature 1kg of substance by [tex]\bold{1^oC}[/tex]. The 54.1 g of water can be heated from [tex]\bold{20^oC}[/tex] to [tex]\bold{ 75^oC}[/tex] using 12500 J.
Specific Heat Capacity:
It is the amount of energy that required to increase the temperature 1kg of substance by [tex]\bold{1^oC}[/tex].
The specific heat formula,
[tex]\bold{ Q = mc \Delta T}[/tex]
Where,
Q - Energy absorbed = 12500 J
m - mass(grams) = ?
c - specific heat = 4,200 J/ kg/[tex]\bold { ^oc}[/tex]
[tex]\bold{ \Delta T}[/tex]- change in temperature in kelvin = 75 -20 = [tex]\bold{55^oC}[/tex]
Put the values is the formula,
[tex]\bold{12500J = m\times 4200J/kg /^oC \times 55^oC}[/tex]
Solve the equation for m
m = 0.0541 kg = 54.1 g
Hence, we can conclude that the 54.1 g of water can be heated from [tex]\bold{20^oC}[/tex] to [tex]\bold{ 75^oC}[/tex] using 12500 J.
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The mass of the water that can be heated from 20 °C to 75 °C using 12500.0 J is 54.32 g
We'll begin by calculating the change in the temperature of the water. This can be obtained as follow:
Initial temperature (T₁) = 20 °C
Final temperature (T₂) = 75 °C
Change in temperature (ΔT) =?
ΔT = T₂ – T₁
ΔT = 75 – 20
ΔT = 55 °C
- Finally, we shall determine the mass of the water. This can be obtained as follow:
Change in temperature (ΔT) = 55 °C
Heat (Q) = 12500.0 J
Specific heat capacity of water (C) = 4.184 J/gºC
Mass of water (M) =?
Q = MCΔT
12500 = M × 4.184 × 55
12500 = M × 230.12
Divide both side by 230.12
M = 12500 / 230.12
M = 54.32 g
Thus, the mass of the water is 54.32 g
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