Given: 25 J of heat, ΔT (change in temperature) = 30°C, c (specific heat capacity of water) = 4.184 J/g°C
Find: Mass of water (m)
Formula: m = Q / (c * ΔT)
Solution: m = 25 J / (4.184 J/g°C * 30°C)
m ≈ 0.188 grams
Answer: The mass of water that will change its temperature by 30°C when 25 J of heat is added to it is approximately 0.188 grams.
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Given: 25 J of heat, ΔT (change in temperature) = 30°C, c (specific heat capacity of water) = 4.184 J/g°C
Find: Mass of water (m)
Formula: m = Q / (c * ΔT)
Solution: m = 25 J / (4.184 J/g°C * 30°C)
m ≈ 0.188 grams
Answer: The mass of water that will change its temperature by 30°C when 25 J of heat is added to it is approximately 0.188 grams.