Answer: Final temperature for given mass of water is [tex]24.6^o C[/tex].
Explanation:
Given: mass = 630.1 g, specific heat = 418 [tex]J/g^{o} C[/tex], q = 950 J, T = [tex]24.2^o C[/tex]
Formula used: [tex]q = mC \Delta T[/tex]
where, q = heat energy
m = mass
C = specific heat
[tex]\Delta T[/tex] = change in temperature = [tex](T_2 - T_1)[/tex]
950 J = [tex]630.1 g \times 4.18 J/g^o C \times (T_2 - 24.2^o)[/tex]
[tex]T_2 = 24.6^o C[/tex]
Answer:
T2 = 24.56°C
Explanation:
Given data:
Mass of water = 630.1 g
Specific heat of water = 4.18 J/g.°C
Initial temperature = 24.2°C
Heat absorbed = 950 J
Final temperature = ?
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
by putting values
950 J = 630.1 g × 4.18 J/g.°C × (T2 - 24.2°C)
950 J = 2633.82 j/°C × (T2 - 24.2°C)
950 J / 2633.82 j/°C = (T2 - 24.2°C)
0.36°C + 24.2°C = T2
T2 = 24.56°C
