jay heated 120mL of water that was 20 degrees celsius to 80 degrees celsius. a)how much did the temperature of water change?
b) how much heat did jay add to the water?
c) if jay used a liquid that was not water, the specific heat would not equal:(circle best answer) 0cal/g/celsius, 1cal/g/celsius, 2cal/g/celsius, or 10cal/g/celsius?
To solve these questions, we will use the formulas related to specific heat and the heat equation.
a) The change in temperature (∆T) is calculated by subtracting the initial temperature (T1) from the final temperature (T2). In this case, T1 = 20°C and T2 = 80°C.
∆T = T2 - T1 = 80°C - 20°C = 60°C
Therefore, the temperature of the water changed by 60°C.
b) The amount of heat (q) added to the water can be calculated using the heat equation:
q = m × c × ∆T
where:
- q represents the heat added to the water
- m is the mass of the water
- c is the specific heat capacity of water
- ∆T is the change in temperature
Since the mass of the water is not given, we cannot calculate the exact amount of heat added to the water. However, we can use the specific heat capacity of water, which is approximately 4.18 J/g°C.
c) The specific heat capacity of a substance determines the amount of heat required to raise the temperature of a unit mass (usually 1 gram) of that substance by 1 degree Celsius.
In this case, since the liquid used by Jay is assumed to be water, the specific heat capacity would equal 1 cal/g/°C. Therefore, you should circle the option: "1 cal/g/°C".