A 10.04-g block of solid tin at 14.77 °C is immersed in a 24.11-g pool of liquid ethylene glycol with a temperature of 63.52 °C. When thermal equilibrium is reached, what is the temperature of the tin and ethylene glycol? Specific heat capacities: tin = 0.213 J/g °C; ethylene glycol = 2.36 J/g °C

61.75 °C

Explanation:

According to the law of conservation of energy,

Qtin = - (Qliquid)

(m×c×∆T)tin = -(m×c×∆T) ethylene glycol

Where; m = mass

c = specific heat capacity

∆T = change in temperature

According to the information given in this question;

For tin: m= 10.04g, c= 0.213 J/g °C, initial temperature= 14.77 °C, final temperature=?

For ethylene glycol: m= 24.11g, c= 2.36 J/g °C, initial temperature= 63.52 °C, final temperature=?

Hence; (m×c×∆T)tin = -(m×c×∆T) ethylene glycol

10.04 × 0.213 × (T - 14.77) = - {24.11 × 2.36 × (T - 63.52)}

2.139 (T - 14.77) = - {56.899 (T - 63.52)}

2.139T - 31.59 = -56.899T + 3614.225

2.139T + 56.899T = 31.59 + 3614.225

59.038T = 3645.815

T = 3645.815/59.038

T = 61.75 °C

The final temperature of tin and ethylene glycol is 61.75 °C

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