How much energy is needed to heat 1 kg of sand, which has a specific heat of 664 j/(kg k) from 30°c to 50°c?

1 Answer
Jun 17, 2014

#1.33×10^4J#


The equation we need to use is: #∆Q=mc∆θ#

The following data is given:
#m=1.0kg#
#c=664 J kg^-1 K^-1#
#∆θ=50-30=20 ºC#

Note that it is not necessary to convert the temperatures into Kelvin as the temperature difference is the same (#323-303=20K#).

Sub. values into the equation:
#∆Q=mc∆θ=1×664×20=1.33×10^4J#