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

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

Answer 2

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

Step-by-step explanation:

`Q=m* c* \Delta T`

`Q=m* c* (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* 10^4 g`

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

Q =
`2.5* 10^3 J`

`2.5* 10^3 J=1.0* 10^4 g* c* (70^oC-10^oC)`

`c=(2.5* 10^3 J)/(1.0* 10^4 g* (70^oC-10^oC))`

`c=0.004167 J/g^oC`

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