Question

A total of 54.0 Joules of heat are absorbed as 58.3 g of lead is heated from 12.0° C to 42.0° C. What is the specific heat of lead? * 1 point 0.031 Joules/grams x degrees Celsius 0.022 Joules/grams x degrees Celsius 0.017 Joules/grams x degrees Celsius 0.029 Joules/grams x degrees Celsius

Answer 1

to increase the temperature of lead we know that the formula of heat required is given as

$Q = ms\Delta T$

here we know that

$Q = 54 J$

m = 58.3 g

$\Delta T = T_f - T_i$

$\Delta T = 42 - 12 = 30 ^0C$

now from the above equation we have

$54 = 0.0583*s *30$

$54 = 1.75*s$

$s = 31 J/kg ^0C$

also we can write its as

$s = 0.031 J/g ^0C$

Answer 2

Answer:

0.031J/kg°C

Explanation:

Quantity of heat absorbed = quantity of heat liberated

Qabsorbed =54J

m= 58.3g

∆T = 42-12= 30°C

Qabsorbed= mc∆T

c = specific heat capacity= Absorbed/m∆T

c = 54/58.3× 30

= 54/1749

c = 0.031J/kg°C