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

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

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Answer 1 · 2016-04-23T16:35:45

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.

$q = m \times C_{s} \times Δ T$ $q = mxxC_sxxDeltaT$

For the equation above:

$q$ $q$ = heat, $m$ $m$ = mass (in grams!), $C_{s}$ $C_s$ = specific heat capacity, and $Δ T$ $DeltaT$ = change in temperature.

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

For sand, $C_{s} = 0.835 \frac{J}{g \times ° C}$ $C_s = 0.835 J/(gxx°C)$

In your specific problem:
$Δ T = T_{f} - T_{i} = 100 - 30 = 70 ° C$ $DeltaT=T_f-T_i= 100 - 30 = 70 °C$
$m = 1.30 k g = 1300 g$ $m = 1.30 kg = 1300 g$
(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:

$q = m \times C_{s} \times Δ T = 1300 \times 0.835 \times 70 =$ $q = mxxC_sxxDeltaT = 1300xx0.835xx70 =$ 79,985 J

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

q = 80,000 J.

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

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