Calculate the heat required to heat 0-5 kg of water from 20 - C to 70 - C- If the specific heat of water is 4200 J-kg-K-

Given: Mass of water, m = 0.5 kg, Specific heat of water, S = 4200 J/kgK Initial temperature, T_1 = 20 ^∘ C Final temperature, T_2 = 70 ^∘ C Hence, temperature ...

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Standard VIII

Physics

Calorimeter

Calculate the...

Question

Calculate the heat required to heat $0.5 k g$ of water from $20^{\circ} C$ to $70^{\circ} C$ . If the specific heat of water is $4200 \frac{J}{k g . K}$ .

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Solution

Given:
Mass of water, $m = 0.5 k g$ ,
Specific heat of water, $S = 4200 \frac{J}{k g K}$
Initial temperature, $T_{1} = 20^{\circ} C$
Final temperature, $T_{2} = 70^{\circ} C$

Hence, temperature change,
$Δ T = T_{2} - T_{1}$
$Δ T = 50^{\circ} C$

Since, amount of heat needed,
$Q = m S Δ T$
$Q = 0.5 \times 4200 \times 50$
$Q = 105, 000 J = 105 k J$

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Calculate the heat required to heat 0.5 kg of water from 20 ∘C to 70 ∘C. If the specific heat of water is 4200 Jkg.K.

Given: Mass of water, m=0.5 kg, Specific heat of water, S=4200 JkgK Initial temperature, T1=20 ∘C Final temperature, T2=70 ∘C Hence, temperature change, ΔT=T2 − T1 ΔT=50 ∘C Since, amount of heat needed, Q=mSΔT Q=0.5×4200×50 Q=105,000 J=105 kJ

Calculate the heat required to heat $0.5 k g$ of water from $20^{\circ} C$ to $70^{\circ} C$ . If the specific heat of water is $4200 \frac{J}{k g . K}$ .