A copper calorimetric cup with a mass of 100g contains 96g of water at 13 C. If 70g of a substance at 84 C is dropped into the cup, the temperature increases to 20 C. Find the specific heat capacity of the substance
6.9x10^2
627
The answer is 6.9 * 10^2 J/kg*K (or swap K with Celsius)
Can someone please explain how you would solve this question? Thank you
To find the specific heat capacity of the substance, we can use the equation:
Q = mcΔT
where Q is the heat gained or lost, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
In this scenario, when the 70g substance at 84°C is dropped into the cup containing 96g of water at 13°C, the final temperature becomes 20°C.
We can break down the heat transfer as follows:
1. Heat lost by the substance: m1c1ΔT1, where m1 is the mass of the substance, c1 is its specific heat capacity, and ΔT1 is the change in temperature of the substance.
2. Heat gained by the water: m2c2ΔT2, where m2 is the mass of the water, c2 is the specific heat capacity of water (which is approximately 4.18 J/g°C), and ΔT2 is the change in temperature of the water.
Since there is no heat exchange with the surroundings, the heat lost by the substance is equal to the heat gained by the water. Therefore, we equate the two expressions:
m1c1ΔT1 = m2c2ΔT2
Plugging in the given values:
70g * c1 * (84°C - 20°C) = 96g * 4.18 J/g°C * (20°C - 13°C)
Simplifying the equation:
70c1 * 64 = 96 * 4.18 * 7
Dividing both sides by 70:
c1 = (96 * 4.18 * 7) / (70 * 64)
Calculating:
c1 ≈ 0.906 J/g°C
Therefore, the specific heat capacity of the substance is approximately 0.906 J/g°C.