Question

A 1.50-kg iron horseshoe initially at 590°C is dropped into a bucket containing 17.0 kg of water at 24.0°C. What is the final temperature of the water–horseshoe system? Ignore the heat capacity of the container and assume a negligible amount of water boils away.

Answer 1

Answer:

29.3  °C

Explanation:

$m_{h}$ = mass of iron horseshoe = 1.50 kg

$T_{h}$ = Temperature of iron horseshoe initially = 590 °C

$c_{h}$ = specific heat of iron = 450 J/(kg °C)

$m_{w}$ = mass of water = 17 kg

$T_{w}$ = Temperature of water initially = 24.0 °C

$c_{w}$ = specific heat of water = 4186 J/(kg °C)

$T_{f}$ = Final equilibrium Temperature

using conservation of heat

Heat lost by iron horseshoe = Heat gained by water

$m_{h} c_{h} (T_{h} - T_{f}) = m_{w} c_{w} (T_{f} - T_{w})$

$(1.50) (450)(590 - T_{f}) = (17) (4186) (T_{f} - 24)$

$T_{f}$ = 29.3  °C