I just got this question and it was lipids sorry if i’m wrong
Answer:
Q = 5671.05 J
Explanation:
Given data:
Mass of metal = 525 g
Initial temperature = 32.0 °C
Fina temperature = 43.0 °C
Heat absorbed by metal = ?
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 metal X = 0.982 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
ΔT = 43.00°C - 32.00°C
ΔT = 11.00°C
Q = 525 g × 0.982 J/g.°C × 11.00°C
Q = 5671.05 J
Answer:
7.74%
<em>Hope this helped! Have a good day.</em>
