Answer:
ΔT = 75.11 °F
Explanation:
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
Given data:
mass= 0.5 g
heat released = 50.1 j
specific heat of water= 4.184 j/g .°C
temperature change =ΔT= ?
Solution:
Q = m × c × ΔT
50.1 j = 0.5 g × 4.184 j/g × ΔT
50.1 j = 2.092 j.°C × ΔT
50.1 j/ 2.092 j.°C = ΔT
23.95°C = ΔT
conversion of °C to °F
(23.95 × 9/5 ) + 32 = 75.11 °F
ΔT = 75.11 °F
