Answer:
1.23 j/g. °C
Explanation:
Given data:
Mass of metal = 35.0 g
Initial temperature = 21 °C
Final temperature = 52°C
Amount of heat absorbed = 320 cal (320 ×4.184 = 1338.88 j)
Specific heat capacity of metal = ?
Solution:
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
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 52°C - 21 °C
ΔT = 31°C
1338.88 j= 35 g ×c× 31°C
1338.88 j= 1085 g.°C ×c
1338.88 j/1085 g.°C = c
1.23 j/g. °C = c
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<span>ions are solutions containing ions that react with acids or bases to minimize their effects. </span>
