How much energy is needed to heat 1 kg of sand, which has a specific heat of 664 j/(kg k) from 30°c to 50°c?

Answer 1
#1.33×10^4J#
The equation we need to use is: #∆Q=mc∆θ#
The following data is given: #m=1.0kg# #c=664 J kg^-1 K^-1# #∆θ=50-30=20 ºC#
Note that it is not necessary to convert the temperatures into Kelvin as the temperature difference is the same (#323-303=20K#).
Sub. values into the equation: #∆Q=mc∆θ=1×664×20=1.33×10^4J#
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Answer 2

The energy needed to heat 1 kg of sand from 30°C to 50°C can be calculated using the formula:

[ Q = mc\Delta T ]

Where:

  • ( Q ) is the energy required (in joules)
  • ( m ) is the mass of the sand (1 kg)
  • ( c ) is the specific heat capacity of the sand (664 J/(kg·K))
  • ( \Delta T ) is the change in temperature (50°C - 30°C = 20°C)

Substituting the values into the formula:

[ Q = (1 kg) \times (664 J/(kg·K)) \times (20°C) ]

[ Q = 13280 J ]

So, the energy needed to heat 1 kg of sand from 30°C to 50°C is 13280 joules.

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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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