Question

what is the specific heat of a substance that absorbs 2.5×10^3 joules of heat when a sample of 1.0 ×10^4g of the substance increases in temperature from 10°c to 70°c?​

Answer 1

Answer:

0.004167 J/g°C is the specific heat of a substance.

Explanation:

$Q=m\times c\times \Delta T$

$Q=m\times c\times (T_{2}-T_1)$

Where:

Q =  heat absorbed(positive) or released (negative)

m = Mass of substance

c =  specific heat of a substance

$T_1$ = Initial temperature

$T_2$ = Final temperature

We have:

m = $1.0\times 10^4 g$

c = ? , $T_1=10^oC,T_2=70^oC$

Q = $2.5\times 10^3 J$

$2.5\times 10^3 J=1.0\times 10^4 g\times c\times (70^oC-10^oC)$

$c=\frac{2.5\times 10^3 J}{1.0\times 10^4 g\times (70^oC-10^oC)}$

$c=0.004167 J/g^oC$

0.004167 J/g°C is the specific heat of a substance.

Answer 2

Sorry I don’t even know what the product would be