Answer:
ΔT = 5.6°C
Explanation:
Given data:
Heat added = 985 j
Mass of water = 42 g
Initial temperature = 50°C
Change in temperature = ?
Solution;
Specific heat capacity:
It is the amount of heat required to raise the temperature of one gram of substance by one degree.
Specific heat capacity of water = 4.18 j/g.°C
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
Now we will put the values.
985 j = 42 g ×4.18 j/g.°C × (Final temperature - initial temperature)
985 j = 42 g ×4.18 j/g.°C × (T2 - 50°C)
985 j = 175.56 j/°C × (T2 -50°C)
985 j / 175.56 j/°C = T2 - 50°C
5.6°C + 50°C = T2
T2 = 55.6°C
Change in temperature = ΔT = Final temperature - initial temperature
ΔT = 55.6°C - 50°C
ΔT = 5.6°C
