An electric heater rated at 200\text{ W} is used to heat a liquid of mass 1.5\text{ kg} fo… — JAMB UTME

An electric heater rated at $200\text{ W}$ is used to heat a liquid of mass $1.5\text{ kg}$ for $7\text{ minutes}$ . If the temperature of the liquid rises from $25^\circ\text{C}$ to $65^\circ\text{C}$ , assuming no heat energy is lost to the surroundings, what is the specific heat capacity of the liquid in $\text{J}\text{kg}^{-1}\text{K}^{-1}$ ?

Answer: 1400 J kg^-1 K^-1

Answer

The specific heat capacity of the liquid is

1400\text{ J}\text{kg}^{-1}\text{K}^{-1}

The electrical energy supplied over

7\text{ minutes}

(

420\text{ s}

) at a power rating of

200\text{ W}

equals

84,000\text{ J}

. Equating this total energy to thermal absorption

Q = m c \Delta T

for a

1.5\text{ kg}

mass experiencing a

40\text{ K}

temperature rise gives

c = \frac{84,000}{1.5 \times 40} = 1400\text{ J}\text{kg}^{-1}\text{K}^{-1}

Step-by-Step Solution

Convert heating time to seconds and compute total heat energy supplied.

t = 7 \times 60 = 420\text{ s}

, so

Q = P \times t = 200 \times 420 = 84,000\text{ J}

Electrical power is the rate of energy transfer (

P = Q/t

), so heat energy equals power multiplied by time in seconds.

Calculate temperature difference.

\Delta T = 65^\circ\text{C} - 25^\circ\text{C} = 40\text{ K}

The temperature rise drives heat absorption according to the thermal equation.

Rearrange the heat energy formula

Q = m c \Delta T

to find the specific heat capacity

c

c = \frac{84,000}{1.5 \times 40} = \frac{84,000}{60} = 1400\text{ J}\text{kg}^{-1}\text{K}^{-1}

Specific heat capacity represents the heat energy required per unit mass per unit temperature change.

Key Concept

Specific heat capacity calculation using electrical energy input (

Q = P t = m c \Delta T

Estimated Time:1m 15s

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