Calculate the calories lost when 95 g of water cools from 45 ∘C to 29 ∘C. Express your answer to two significant figures and include the appropriate units. Use abbreviations for the units.

Question

Myles Ferguson

Physics

23 July, 19:58

Calculate the calories lost when 95 g of water cools from 45 ∘C to 29 ∘C. Express your answer to two significant figures and include the appropriate units. Use abbreviations for the units.

+4

Answers (1)

Know the Answer?

Robert · Answer 1

1,520.00 calories

Explanation:

Water molecules are linked by hydrogen bonds that require a lot of heat (energy) to break, which is released when the temperature drops. That energy is called specific heat or thermal capacity (ĉ) when it is enough to change the temperature of 1g of the substance (in this case water) by 1°C. Water ĉ equals 1 cal / (g.°C).

Given that ĉ = Q / (m.ΔT),

where Q = calories transferred between the system and its environment or another system (unity: calorie or cal) (what we are trying to find out),

m = mass of the substance (unity: grams or g), and

ΔT = difference of temperature (unity: Celsius degrees or °C); and

m = 95g and ΔT = 16°C:

Q = 1 cal / (g.°C).95g. 16°C = 1,520.00 cal