Question

How much heat is required to warm 1.30 kg of sand from 30°C to 100°C?

Answer 1

Heating 1.30 Kg of sand 70 degrees Celsius takes 80,000 Joules of heat.

Explanation:

The amount of heat required to warm a substance is determined using heat capacity. The equation below allows us to solve for the heat released or absorbed when warming a substance.

`q = mxxC_sxxDeltaT`

For the equation above:

`q` = heat, `m` = mass (in grams!), `C_s` = specific heat capacity, and `DeltaT`= change in temperature.

Specific heat capacity (`C_s`) tells us how hard or easy it is to heat up a substance. You can look up the `C_s` for different substances in tables in your textbook or on the internet.

For sand, `C_s = 0.835 J/(gxx°C)`

In your specific problem:
`DeltaT=T_f-T_i= 100 - 30 = 70 °C`
`m = 1.30 kg = 1300 g`
(you must convert to grams when using specific heat capacities since they are in the units of grams)

Next we plug our values into the top equation listed:

`q = mxxC_sxxDeltaT = 1300xx0.835xx70 =` 79,985 J

Since your temperature has only one significant figure we round to one sig fig:

q = 80,000 J.

For more examples see this video: