A68-g sample of sodium is at an initial temperature of 42 °c. if 1840. joules of heat are applied to the sample, what is the final temperature of the sodium?

Final T = 64.0°C.

Explanation:

Q = m.c.ΔT.

where, Q is the amount of heat absorbed by Na (Q = 1840 J),

m is the mass of Na (m = 68.0 g),

c is the specific heat capacity of Na (c = 1.23 J/g °C),

ΔT is the temperature difference (final T - initial T) (ΔT = final T - 42.0°C).

∵ Q = m.c.ΔT.

∴ (1840 J) = (68.0 g)(1.23 J/g °C)(final T - 42.0°C)

(final T - 42.0°C) = (1840 J)/(68.0 g)(1.23 J/g °C) = 22.0°C.

∴ final T = 22.0°C + 42.0°C = 64.0°C.

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