Question

A hot iron horseshoe (mass = 0.35 kg ), just forged, is dropped into 1.40 L of water in a 0.45 kg iron pot initially at 22.0°C. Part A If the final equilibrium temperature is 32.0°C, estimate the initial temperature of the hot horseshoe. Express your answer using two significant figures.

Answer 1

Answer:

$420$ °C

Explanation:

$m_{h}$ = mass of the horseshoe = 0.35 kg

$m_{w}$ = mass of the water = 1.40 L = 1.40 kg

$m_{i}$ = mass of the iron pot = 0.45 kg

$c_{i}$ = specific heat of iron = 450 J kg⁻¹ °C⁻¹

$c_{w}$ = specific heat of water = 4186 J kg⁻¹ °C⁻¹

$T_{hi}$ = initial temperature of the horseshoe = ?

$T_{wi}$ = initial temperature of the water = 22 °C

$T_{pi}$ = initial temperature of the iron pot = 22 °C

$T_{f}$ = final temperature = 32 °C

Using conservation of Heat

$m_{h}c_{i}(T_{hi} - T_{f}) = m_{w}c_{w}(T_{f} - T_{wi}) + m_{i}c_{i}(T_{f} - T_{pi})$

$(0.35)(450)(T_{hi} - 32) = (1.40)(4186)(32 - 22) + (0.45)(450)(32 - 22)$

$(157.5)(T_{hi} - 32) = 60629$

$T_{hi} - 32 = 384.95$

$T_{hi} = 420$ °C