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

Apr 23, 2016

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.

$q = m \times {C}_{s} \times \Delta T$

For the equation above:

$q$ = heat, $m$ = mass (in grams!), ${C}_{s}$ = specific heat capacity, and $\Delta T$= 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)

DeltaT=T_f-T_i= 100 - 30 = 70 °C
$m = 1.30 k g = 1300 g$
(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:

$q = m \times {C}_{s} \times \Delta T = 1300 \times 0.835 \times 70 =$ 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: