Question

A \[10\,{\text{watt}}\] electric heater is used to heat a water container filled with \[0.5\,{\text{kg}}\] of water. It is found that the temperature of the water and the container rises by \[3^\circ {\text{K}}\] in 15 minutes. The container is then emptied, dried and filled with \[2\,{\text{kg}}\] of oil. The same heater now raises the temperature of the container-oil system by \[2^\circ {\text{K}}\] in 20 minutes. Assuming that there is no heat loss in the process and the specific heat of water is \[4200\,{\text{J}} \cdot {\text{k}}{{\text{g}}^{ - 1}} \cdot {{\text{K}}^{ - 1}}\], the specific heat of oil in the same unit is equal to
A. \[1.50 \times {10^3}\]
B. \[2.55 \times {10^3}\]
C. \[3.00 \times {10^3}\]
D. \[5.10 \times {10^3}\]

Answer 1

Hint: Use the formula for power in terms of the heat energy and time. Also use the formula for the heat exchanged by the substance in terms of mass of the substance, specific heat of the substance and change in temperature of the substance. Use the law of conservation of energy to the heater and the system of container-water and container-oil.

Formulae used:
The heat \[Q\] exchanged by a substance is
\[Q = mc\Delta T\] …… (1)
Here, \[m\] is the mass of the substance, \[c\] is specific heat of the substance and \[\Delta T\] is the change in temperature of the substance.
The power \[P\] is given by
\[P = \dfrac{Q}{t}\] …… (2)
Here, \[Q\] is the heat energy and \[t\] is the time.

Complete step by step answer:
We have given that the power of the heater is \[10\,{\text{W}}\].
\[P = 10\,{\text{W}}\]
We have given that the mass of the water in the container is \[0.5\,{\text{kg}}\] and the temperature of the container-water system rises by \[3^\circ {\text{K}}\] in 15 minutes.
\[{m_W} = 0.5\,{\text{kg}}\]
\[\Delta {T_W} = 3^\circ {\text{K}}\]
\[{t_W} = 15\min \]
The specific heat of water is \[4200\,{\text{J}} \cdot {\text{k}}{{\text{g}}^{ - 1}} \cdot {{\text{K}}^{ - 1}}\].
\[{c_W} = 4200\,{\text{J}} \cdot {\text{k}}{{\text{g}}^{ - 1}} \cdot {{\text{K}}^{ - 1}}\]
We have asked to calculate the specific heat of oil.We have also given that there is no loss of heat during the process of heating of the water. Hence, according to the law of conservation of energy, the heat energy supplied by the heater to the water-container system is equal to the heat gained by the water and container.
\[P{t_W} = {m_W}{c_W}\Delta {T_W} + {m_C}{c_C}\Delta {T_W}\]
\[P{t_W} = \left( {{m_W}{c_W} + {m_C}{c_C}} \right)\Delta {T_W}\]
Substitute \[10\,{\text{W}}\] for \[P\], \[15\min \] for \[{t_W}\], \[0.5\,{\text{kg}}\] for \[{m_W}\], \[4200\,{\text{J}} \cdot {\text{k}}{{\text{g}}^{ - 1}} \cdot {{\text{K}}^{ - 1}}\] for \[{c_W}\] and \[3^\circ {\text{K}}\] for \[\Delta {T_W}\] in the above equation.
\[\left( {10\,{\text{W}}} \right)\left( {15\min } \right) = \left[ {\left( {0.5\,{\text{kg}}} \right)\left( {4200\,{\text{J}} \cdot {\text{k}}{{\text{g}}^{ - 1}} \cdot {{\text{K}}^{ - 1}}} \right) + {m_C}{c_C}} \right]\left( {3^\circ {\text{K}}} \right)\]
\[ \Rightarrow \left( {10\,{\text{W}}} \right)\left[ {\left( {15\min } \right)\left( {\dfrac{{60\,{\text{s}}}}{{1\,{\text{min}}}}} \right)} \right] = \left[ {\left( {0.5\,{\text{kg}}} \right)\left( {4200\,{\text{J}} \cdot {\text{k}}{{\text{g}}^{ - 1}} \cdot {{\text{K}}^{ - 1}}} \right) + {m_C}{c_C}} \right]\left( {3^\circ {\text{K}}} \right)\]
\[ \Rightarrow 9000 = 6300 + 3{m_C}{c_C}\]
\[ \Rightarrow 3{m_C}{c_C} = 9000 - 6300\]
\[ \Rightarrow 3{m_C}{c_C} = 2700\]
\[ \Rightarrow {m_C}{c_C} = 900\,{\text{J}} \cdot {{\text{K}}^{ - 1}}\]

Now we have given that the mass of the oil added to the container is \[2\,{\text{kg}}\] and the temperature of the container-oil system rises by \[2^\circ {\text{K}}\] in 20 minutes.
\[{m_O} = 2\,{\text{kg}}\]
\[\Delta {T_O} = 2^\circ {\text{K}}\]
\[{t_O} = 20\min \]
According to the law of conservation of energy, the heat energy supplied by the heater to the oil-container system is equal to the heat gained by the oil and container.
\[P{t_O} = {m_O}{c_O}\Delta {T_O} + {m_C}{c_C}\Delta {T_O}\]
\[ \Rightarrow P{t_O} = \left( {{m_O}{c_O} + {m_C}{c_C}} \right)\Delta {T_O}\]
Substitute \[10\,{\text{W}}\] for \[P\], \[20\min \] for \[{t_O}\], \[2\,{\text{kg}}\] for \[{m_W}\], \[900\,{\text{J}} \cdot {{\text{K}}^{ - 1}}\] for \[{m_C}{c_C}\] and \[2^\circ {\text{K}}\] for \[\Delta {T_O}\] in the above equation.
\[ \Rightarrow \left( {10\,{\text{W}}} \right)\left( {20\min } \right) = \left[ {\left( {2\,{\text{kg}}} \right){c_O} + \left( {900\,{\text{J}} \cdot {{\text{K}}^{ - 1}}} \right)} \right]\left( {2^\circ {\text{K}}} \right)\]
\[ \Rightarrow \left( {10\,{\text{W}}} \right)\left[ {\left( {20\min } \right)\left( {\dfrac{{60\,{\text{s}}}}{{1\,{\text{min}}}}} \right)} \right] = \left[ {\left( {2\,{\text{kg}}} \right){c_O} + \left( {900\,{\text{J}} \cdot {{\text{K}}^{ - 1}}} \right)} \right]\left( {2^\circ {\text{K}}} \right)\]
\[ \Rightarrow 12000 = 4{c_O} + 1800\]
\[ \Rightarrow 4{c_O} = 12000 - 1800\]
\[ \Rightarrow 4{c_O} = 10200\]
\[ \Rightarrow {c_O} = \dfrac{{10200}}{4}\]
\[ \Rightarrow {c_O} = 2550\,{\text{J}} \cdot {\text{k}}{{\text{g}}^{ - 1}} \cdot {{\text{K}}^{ - 1}}\]
\[ \therefore {c_O} = 2.55 \times {10^3}\,{\text{J}} \cdot {\text{k}}{{\text{g}}^{ - 1}} \cdot {{\text{K}}^{ - 1}}\]
Therefore, the specific heat of the oil is \[2.55 \times {10^3}\,{\text{J}} \cdot {\text{k}}{{\text{g}}^{ - 1}} \cdot {{\text{K}}^{ - 1}}\].

Hence, the correct option is B.

Note:The students should keep in mind that the power supplied by the heater for both the cases of the container-water and container-oil system is the same. The only change is in the time required for the system to attain a given temperature change for the cases of the container-water and container-oil system.