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Mashutka [201]
2 years ago
11

100. g of ice at 0 is added to 300.0 g of water at 60C. Assuming no transfer of heat to

Chemistry
1 answer:
ololo11 [35]2 years ago
8 0
Heat of fusion of water = 333 J/g 100 g ice x 333 J/g = 33300 J (energy to melt 100 g ice at 0 deg C) 300
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Calculate the specific heat capacity for a 15.3-g sample of gold that absorbs 87.2 J when its temperature increases from 35.0 °C
diamong [38]

Answer:

The specific heat of gold is 0.129 J/g°C

Explanation:

Step 1: Data given

Mass of gold  = 15.3 grams

Heat absorbed = 87.2 J

Initial temperature = 35.0 °C

Final temperature = 79.2 °C

Step 2:

Q = m*c*ΔT

⇒ Q =the heat absorbed = 87.2 J

⇒ m = the mass of gold = 15.3 grams

⇒ c = the specific heat of gold = TO BE DETERMINED

⇒ ΔT = The change in temperature = T2 - T1 = 79.2 - 35.0 = 44.2 °C

87.2 J = 15.3g * c * 44.2°C

c = 87.2 / (15.3 * 44.2)

c = 0.129 J/g°C

The specific heat of gold is 0.129 J/g°C

4 0
3 years ago
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