Question

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`

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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`