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

1 Answer
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 = 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: