Answer:
T(final) = 328.78 K
Explanation:
Given data:
Energy added = 980 kj
Mass of water = 6.2 L
Initial temperature = 291 K
Final temperature = ?
Solution:
Mass of water = 6.2 L = 6.2 Kg = 6200 g
Specific heat of water = 4.184 j/g.K
Energy added = 980 kj = 980 × 1000 = 980,000 J
Specific heat capacity:
It is the amount of heat required to raise the temperature of one gram of substance by one degree.
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
980,000 J = 6200 g × 4.184 j/g.K × T(final) - 291 K
980,000 J = 25940.8 j/K × T(final) - 291 K
980,000 J / 25940.8 j/K = T(final) - 291 K
T(final) - 291 K = 37.78 K
T(final) = 37.78 K + 291 K
T(final) = 328.78 K
