The specific heat capacity of platinum is 0.032 cal/g·K. How to calculate the heat, in joules, necessary to raise the temperature of a sample of platinum weighing 20.0 g from 15.0 Celsius to 65.0 Celsius?

Answer 1

You have to convert either the specific heat of platinum, or the actual heat required from Cal/g K to J/g K. I"ll convert the specific heat and calculate the heat directly in Joules.

So, in order to get from calories per g Kelvin to Joules per g Kelvin you must use a conversion factor

#0.032"cal"/("g" * "K") * "4.184 J"/"1 calorie" = 0.134"J"/("g" * "K")#

The relationship between heat and temperature change is expressed mathematically through this equation

#q = m * c * DeltaT#, where

#q# - the heat supplied;
#m# - the mass of the sample;
#c# - the specific heat of, in your case, platinum;
#DeltaT# - the change in temperature;

Solving for #q# will get you

#q = "20.0 g" * 0.134"J"/("g" * "K") * (338.15 - 288.15)"K" = "+134 J"#

Notice how little energy is needed to raise the temperature of a #"20.0-g"# sample of platinum by 50 degrees; by comparison, you would need approximately 30 times more energy to raise the temperature of this much water by 50 degrees.

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Answer 2

To calculate the heat necessary, you would use the formula:

Q = m * c * ΔT

Where:

  • Q is the heat (in joules)
  • m is the mass of the substance (in grams)
  • c is the specific heat capacity (in cal/g·K)
  • ΔT is the change in temperature (in Celsius)

Plugging in the given values: m = 20.0 g c = 0.032 cal/g·K ΔT = (65.0°C - 15.0°C) = 50.0°C

Q = 20.0 g * 0.032 cal/g·K * 50.0°C

Convert cal to joules: 1 cal = 4.184 J

Q = (20.0 g * 0.032 cal/g·K * 50.0°C) * 4.184 J/cal

Calculate Q.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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