Question

If 1.22 kJ of heat is added to 60.0 g of water at 29.0 ◦C, what is the final temperature of the water? (The specific heat of water is 4.184 J/g ◦C)

An unknown metal with a mass of 36.0 g absorbs 58 J of heat. Its temperature rises from 10.4 ◦C to 12.9 ◦C. What is the specific heat of the unknown metal?

Answer 1

(1)$\; ∆H = mC(T_2-T_1)$

$m = 60 \;g$

$C = 4.184 \;J/g ◦C$

$T_1 = 29.0\; ◦C$

$∆H = 1220\; J$

$1220 = 60\times 4.184\times(T_2 – 29.0)$

$4.86 = T_2 – 29.0$

$T_2 = 33.86 \;◦C$

(2) $∆H = mC(T_2-T_1)$

$m = 36.0 \;g$

$T_1 = 10.4\; ◦C$

$T_2 = 12.9\; ◦C$

$∆H = 58 \;J$

$58 = 36.0\times C\times (12.9-10.4)$

$58 = 90\;C$

$C = 0.64 \;J/g ◦C$