# The Calorimeter Mystery

![Figure 1: Clean labelled diagram in the style of a textbook figure illustrating Enthalpy changes and calorimetry: the equipment, materials and key](https://goa-cc-uat-aili-app-001.azurewebsites.net/api/generate/b543fb5b-e5a2-49ee-b0a2-ab3ba2c0f32b/asset/931)

In Chemistry 30, you apply the relationship Q = mcΔt to analyze heat transfer in calorimetry experiments ([Outcome](https://goa-cc-uat-aili-app-001.azurewebsites.net/explore/node/23634)). You also use calorimetry data to determine enthalpy changes in chemical reactions ([Outcome](https://goa-cc-uat-aili-app-001.azurewebsites.net/explore/node/23641)) and interpret ΔH notation to communicate energy changes ([Outcome](https://goa-cc-uat-aili-app-001.azurewebsites.net/explore/node/23638)). This puzzle challenges you to work backward from calorimetry results to find a missing piece of experimental data.

## The puzzle

A student burns a food sample in a bomb calorimeter. The calorimeter contains 500 grams of water. After the combustion, the water temperature rises from 22°C to 28°C. The student records the heat released by the food as 12,600 joules.

However, the student forgot to record one crucial measurement: the specific heat capacity of water used in the calculation. The student knows that the specific heat capacity of pure water is 4.18 J/(g·°C), but wonders whether the value actually used in the Q = mcΔt equation matches this standard value.

Calculate what specific heat capacity value the student must have used in the Q = mcΔt equation to arrive at the recorded result of 12,600 joules. Show your reasoning. Does this value match the standard specific heat capacity of water?

## Hints

**Hint 1:** Rearrange Q = mcΔt to solve for the unknown variable. What are you solving for?

**Hint 2:** You know Q (heat released), m (mass of water), and ΔT (temperature change). The only unknown is c, the specific heat capacity. Substitute the numbers you have and solve for c.

**Hint 3:** Calculate ΔT first: 28°C − 22°C = 6°C. Then divide Q by the product of m and ΔT to find c. Your answer should be in units of J/(g·°C).

## Answer key

**The answer:** The specific heat capacity used was approximately 4.2 J/(g·°C).

**The reasoning:**

Start with Q = mcΔt and rearrange to solve for c:

c = Q ÷ (m × ΔT)

Substitute the known values:

c = 12,600 J ÷ (500 g × 6°C)

c = 12,600 J ÷ 3,000 g·°C

c = 4.2 J/(g·°C)

This value is very close to the standard specific heat capacity of pure water, 4.18 J/(g·°C). The small difference (4.2 versus 4.18) results from the rounded values chosen for this puzzle. The student's calculation is sound and consistent with the accepted value for water.

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*AILI game · language en · model claude-haiku-4-5-20251001 · generated 2026-09-17 · id b543fb5b-e5a2-49ee-b0a2-ab3ba2c0f32b*

### Sources

- node:n1: Chemistry › Chemistry (20, 30) › Chemistry 30 › Unit A: Thermochemical Changes › General Outcome 1 (https://goa-cc-uat-aili-app-001.azurewebsites.net/explore/node/23636)
- node:n2: Chemistry › Chemistry (20, 30) › Chemistry 30 › Unit A: Thermochemical Changes › General Outcome 1 (https://goa-cc-uat-aili-app-001.azurewebsites.net/explore/node/23634)
- node:n3: Chemistry › Chemistry (20, 30) › Chemistry 30 › Unit A: Thermochemical Changes › General Outcome 1 (https://goa-cc-uat-aili-app-001.azurewebsites.net/explore/node/23641)
- node:n4: Chemistry › Chemistry (20, 30) › Chemistry 30 › Unit A: Thermochemical Changes › General Outcome 1 (https://goa-cc-uat-aili-app-001.azurewebsites.net/explore/node/23634)
- node:n5: Chemistry › Chemistry (20, 30) › Chemistry 30 › Unit A: Thermochemical Changes › General Outcome 1 (https://goa-cc-uat-aili-app-001.azurewebsites.net/explore/node/23636)
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- node:n7: Chemistry › Chemistry (20, 30) › Chemistry 30 › Unit A: Thermochemical Changes (https://goa-cc-uat-aili-app-001.azurewebsites.net/explore/node/23633)
- node:n8: Chemistry › Chemistry (20, 30) › Chemistry 30 › Unit A: Thermochemical Changes › General Outcome 1 (https://goa-cc-uat-aili-app-001.azurewebsites.net/explore/node/23648)
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