Answer:
Specific heat of calcium carbonate(C) = 0.82 (Approx)
Explanation:
Given:
Energy absorbs (q) = 85 J
Change in temperature (Δt) = 34.9 - 21 = 13.9°C
Mass of calcium carbonate = 7.47 g
Find:
Specific heat of calcium carbonate(C)
Computation:
Specific heat of calcium carbonate(C) = q / m(Δt)
Specific heat of calcium carbonate(C) = 85 / (7.47)(13.9)
Specific heat of calcium carbonate(C) = 85 / 103.833
Specific heat of calcium carbonate(C) = 0.8186
Specific heat of calcium carbonate(C) = 0.82 (Approx)
Answer:
10425 J are required
Explanation:
assuming that the water is entirely at liquid state at the beginning , the amount required is
Q= m*c*(T final - T initial)
where
m= mass of water = 25 g
T final = final temperature of water = 100°C
T initial= initial temperature of water = 0°C
c= specific heat capacities of water = 1 cal /g°C= 4.186 J/g°C ( we assume that is constant during the entire temperature range)
Q= heat required
therefore
Q= m*c*(T final - T initial)= 25 g * 4.186 J/g°C * (100°C- 0°C) = 10425 J
thus 10425 J are required
Hi,
The answer should be C.
Hope this helps, if you’d like further explanation please let me know.