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Question: How much heat is required to warm \[1.30kg\] of sand from \[{22^ \circ }C\] to \[{100^ \circ }C\] ?...

How much heat is required to warm 1.30kg1.30kg of sand from 22C{22^ \circ }C to 100C{100^ \circ }C ?

Explanation

Solution

Here our duty is to find the heat required to warm the sand. For finding a particular heat released or absorbed when warming a substance , we need to proceed through a specific formula which is given below in the formula used which is the equation for heat capacity. To follow the equation we should know the specific heat capacity of sand. Actually specific heat capacity Cs{C_s} determines how hard or easy it is to heat up a substance. It will be different for different substances.

Complete answer:

Actually here we should know what heat capacity is before moving to a solution. Heat capacity determines the amount of heat which is required to warm a substance . The particular equation for heat capacity is mentioned above in the formula used.

Next, we shall move onto the solution.

From the equation in the formula used, we actually know the mass as given in the question which is 1.30kg1.30kg. But here we need to convert it to grams since specific heat capacities are denoted in units of grams. Therefore the mass, m=1.30kg=1300gm = 1.30kg = 1300g

Next, we have to look into Cs{C_s} which is the specific heat capacity of sand that should be known to us by referring to various sources and the value is 0.8350.835 . For sand , here we actually write Cs{C_s}as following,

Cs=0.835×Jg×C{C_s} = 0.835 \times \dfrac{J}{{g \times {}^ \circ C}}

Next, we have to look at ΔT\Delta T on the equation, which is actually the temperature difference between final temperature and initial temperature. Here, ΔT\Delta T is as follows,

ΔT=TfTi=10030=70C\Delta T = {T_f} - {T_i} = 100 - 30 = {70^ \circ }C

Next , our duty is to substitute the above values in the top equation as mentioned in the formula used.

q=m×Cs×ΔTq = m \times {C_s} \times \Delta T

q=1300×0.835×70q = 1300 \times 0.835 \times 70

q=75985J q= 75985J

We know the temperature difference we find above has only one significant figure and therefore we need to round this final answer to one significant figure also . So , the final answer will be as follows,

q=76000Jq = 76000J

Hence, the heat required to warm 1.30kg1.30kg of sand from 22C{22^ \circ }C to 100C{100^ \circ }C is 76000J76000J.

Note: Here, as in this particular question there will be a great chance of getting confused between heat capacity and specific heat capacity. Actually both are different as heat capacity will be greatly varied according to temperature and other factors whereas specific heat capacity is fixed for a particular substance which varies from substance to substance. We can look up the specific heat capacity of different substances from textbooks , the internet or any other source. We also need to be concerned in converting from kilograms to grams of mass given as it is very important since specific heat capacities are in units of grams.