Question

A 70.0-g piece of metal at 80.0 °C is placed in 100 g of water at 22.0 °C contained in a calorimeter like that shown in Figure 5.12. The metal and water come to the same temperature at 24.6 °C. How much heat did the metal give up to the water? What is the specific heat of the metal?

Short answer

The heat given by the metal to the water = 1088J.

The specific heat of the metal = \(0.280\frac{J}{{{g^0}C}}\).

Step-by-step solution

Net heat change     

Under ideal circumstances, the net heat change is zero.

\({q_{subs\tan ceM}} + {q_{subs\tan ceN}}\) = 0

This relationship can be rearranged to show that the heat gained by the substance

M is equal to the heat lost by the substance W.

\({q_{subs\tan ceM}} = - {q_{subs\tan ceN}}\)

The negative sign merely shows that the direction of the heat flow is opposite to each other.

Calculation of heat

Given information:

The mass of the water = 100 g

The change in the temperature of the water =\({T_{final}} - {T_{initial}} = \)(24.6–22)℃= 2.6℃

The specific heat of the water = 4.186\(\frac{J}{{{g^0}C}}\)

Therefore, Q = \(100 \times 4.186 \times 2.6 = 1088J\)

Heat given by the metal

We know that\({{\bf{q}}_{{\bf{substanceM}}}}{\bf{ = - }}{{\bf{q}}_{{\bf{substanceN}}}}\).

The heat given by the metal to the water is 1088 J.

Specific heat of the metal     

The mass of the metal = 70g.

The change in the temperature of the metal =\({T_{final}} - {T_{initial}} = \)(80–24.6)℃= 55.4℃

Therefore,

1088 J =\(70 \times C \times 55.4\)

\({C_{metal}} = \frac{{1088}}{{70 \times 55.4}} = 0.280\frac{J}{{{g^0}C}}\).

The specific heat of the metal = \(0.280\frac{J}{{{g^0}C}}\).