Calculate the energy required to heat of 1.50 kg iron from -7.8 c to 15.0 c . assume the specific heat capacity of silver under these conditions is .0235 j*g^-1*k^-1 . be sure your answer has the correct number of significant digits.

There is 15.4 kJ of energy required to heat 1.50 kg iron.

Explanation:

Step 1: Data given

Mass of iron = 1.50 kg = 1500 grams

Initial temperature iron = -7.8 °C

Final temperature iron = 15.0 °C

Specific heat capacity of iron = 0.450 J/g*K = 0.450 J/g°C

Note: when heating iron from -7.8 °C to 15.0 °C, the iron does not change to another phase.

Step 2: Calculate the energy required

q = m*c*ΔT

⇒ with Q = the heat transfer ( in Joules)

⇒ with m= the mass of iron = 1.5 kg = 1500 grams

⇒ with c= the specific heat capacity of ir = 0.450 J/g°C

⇒ with ΔT = The change in temperature = T2 - T1 = 15.0 + 7.8 = 22.8 °C

q = 1500 * 22.8 * 0.450

q = 15390 J = 15.4 kJ

This is an endothermic reaction. There is 15.4 kJ of energy required to heat 1.50 kg iron.

in theory, limestone is composed principally of the compound calcium carbonate - caco3, but in reality this compound is deposited in conditions involving a number of other compounds as impurities so it ends up being a mixture of many compounds.

hope this : -)

tony (lordcaos066)

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml