Answer:
ΔT = 0.843 °C
Explanation:
Given data:
Heat added = 0.484 J
Mass of water = 0.1372 g
Temperature increase = ΔT = ?
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
Specific heat capacity of water is 4.184 j/g.°C
Now we will put the values in formula.
0.484 j = 0.1372 g × 4.184 j/g.°C× ΔT
0.484 j = 0.574j/°C× ΔT
ΔT = 0.484 j / 0.574j/°C
ΔT = 0.843 °C
