Ethylene glycol has a specific heat of 0.578 cal/(g•°C). If the temperature of a 23.2 g sample of ethylene glycol is raised from 10.00 °C to 15.64 °C, how much heat energy is absorbed by the sample?
Question
Ethylene glycol has a specific heat of 0.578 cal/(g•°C). If the temperature of a 23.2 g sample of ethylene glycol is raised from 10.00 °C to 15.64 °C, how much heat energy is absorbed by the sample?
Solution 1
To solve this problem, we can use the formula for heat transfer:
q = mcΔT
where:
- q is the heat energy absorbed or released,
- m is the mass of the substance,
- c is the specific heat of the substance, and
- ΔT is the change in temperature.
Given:
- m = 23.2 g,
- c = 0.578 cal/(g•°C), and
- ΔT = 15.64 °C - 10.00 °C = 5.64 °C.
Substituting these values into the formula, we get:
q = (23.2 g)(0.578 cal/(g•°C))(5.64 °C) = 75.2 cal.
So, the sample of ethylene glycol absorbs 75.2 calories of heat energy.
Solution 2
To solve this problem, we can use the formula for heat transfer:
q = mcΔT
where:
- q is the heat energy absorbed or released,
- m is the mass of the substance,
- c is the specific heat of the substance, and
- ΔT is the change in temperature.
Given:
- m = 23.2 g,
- c = 0.578 cal/(g•°C), and
- ΔT = 15.64 °C - 10.00 °C = 5.64 °C.
Substituting these values into the formula, we get:
q = (23.2 g)(0.578 cal/(g•°C))(5.64 °C) = 75.2 cal.
So, the sample of ethylene glycol absorbs 75.2 calories of heat energy.
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