Answer: θ2 = 50.93°
Step-by-step explanation:
Given the following :
Quantity of heat = 75kj = 75 × 1000 = 75000J
Mass of liquid water = 1200g
Temperature of water = 36°C
Using the relation :
Q = mc∆T ; where
Q = heat energy (Joules, J)
m = mass of a substance (kg)
c = specific heat (units J/g.°C)
∆T = change in temperature (θ2 - θ1)
(The specific heat capacity of liquid water(c)
is 4.186 J/g.°C
Inputting values:
Q = mc(θ2 - θ1)
75000 = 1200 × 4.186 (θ2 - 36)
75000 = 1200 × 4.186(θ2) - 150.696
75000 = 5023.2(θ2) - 180835.2
75000 + 180835.2 = 5023.2θ2
255835.2 = 5023.2(θ2)
θ2 = 255835.2 / 5023.2
θ2 = 50.93°