Answer:
368.5 J/Kg℃.
Step-by-step explanation:
The specific heat capacity of a substance is the amount of heat required to raise the temperature of 1 gram of that substance by 1 degree Celsius.
In this problem, we are given the following information:
- Mass of the unknown solid substance: 16 kg
- Initial temperature of the unknown solid substance: 95℃
- Mass of the water: 25 kg
- Initial temperature of the water: 20℃
- Final temperature of the mixture: 24℃
- Specific heat capacity of water: 4186 J/Kg℃
We can use the following equation to calculate the specific heat capacity of the unknown substance:
Q = mcΔT
where:
- Q is the heat energy transferred
- m is the mass of the substance
- c is the specific heat capacity of the substance
- ΔT is the change in temperature
We can calculate the heat energy transferred as follows:
Q = m * c * ΔT
Q = 16 kg * c * (95℃ - 24℃)
Q = 16 kg * c * 71℃
We know that the heat energy transferred is equal to the heat energy lost by the water. The heat energy lost by the water can be calculated as follows:
Q = m * c * ΔT
Q = 25 kg * 4186 J/Kg℃ * (24℃ - 20℃)
Q = 25 kg * 4186 J/Kg℃ * 4℃
Q = 418600 J
Since the heat energy transferred is equal to the heat energy lost, we can set the two equations equal to each other and solve for c:
16 kg * c * 71℃ = 418600 J
c = 418600 J / (16 kg * 71℃)
c = 368.5 J/Kg℃
Therefore, the specific heat capacity of the unknown substance is 368.5 J/Kg℃.